Sure, I'd be happy to help you with completing the square!

To complete the square for a quadratic expression of the form ax2+bx+c" role="presentation" style="position: relative;">ax2+bx+cax2+bx+cax^2+bx+c, follow these steps:ax2+bx+c" role="presentation" style="position: relative;">ax2+bx+cax2+bx+cax^2+bx+c, follow these

1. a" role="presentation" style="position: relative;">aaa (if $a\neqDivide both sides of the equation by a" role="presentation" style="position: relative;">aaa (if a&#x2260;1" role="presentation" style="position: relative;">a≠1a≠1a\neq 1) to get the coefficient of the squared term to be 1.
2. Move the constant term c" role="presentation">ccc to the right-hand side of the equation.c" role="presentation" style="position: relative;">ccc to the right-hand
3. Add and subtract (b/2a)2" role="presentation" style="position: relative;">(b/2a)2(b/2a)2(b/2a)^2 on the right-hand side of the equation. This step is what "completes the square," as it creates a perfect square trinomial on the right-hand side.(b/2a)2" role="presentation" style="position: relative;">(b/2a)2(b/2a)2(b/2a)^2 on the right-hand side of the equation. This step is what "completes the square," as it creates a perfect square trinomial on the right-hand(b/2a)2" role="presentation">(b/2a)2(b/2a)2(b/2a)^2 on the right-hand side of the equation.
4. Simplify the right-hand side of the equation by factoring the perfect square trinomial.
5. Take the square root of both sides of the equation and solve for x" role="presentation" style="position: relative;">xxx.

Here's an example to illustrate the process. Suppose we want to complete the square for the quadratic expression 2x2+8x+5" role="presentation" style="position: relative;">2x2+8x+52x2+8x+52x^2 + 8x + 5. We would proceed as follows:

1. 12(2x2+8x+5)=12(0)" role="presentation" style="text-align: center; position: relative;">12(2x2+8x+5)=12(0)12(2x2+8x+5)=12(0)\frac{1}{2}(2x^2+8x+5) = \frac{1}{2}(0)
This simplifies to: $$x^2 + 4x + \frac{12(2x2+8x+5)=12(0)" role="presentation" style="text-align: center; position: relative;">12(2x2+8x+5)=12(0)12(2x2+8x+5)=12(0)\frac{1}{2}(2x^2+8x+5) = \frac{1}{2}(0)
This simplifies toDivide both sides of the equation by 2: 12(2x2+8x+5)=12(0)" role="presentation">12(2x2+8x+5)=12(0)12(2x2+8x+5)=12(0)\frac{1}{2}(2x^2+8x+5) = \frac{1}{2}(0)
This simplifies to: x2+4x+52=0" role="presentation">x2+4x+52=0x2+4x+52=0x^2 + 4x + \frac{5}{2} = 0
2. Move the constant term 52" role="presentation">5252\frac{5}{2} to the right-hand side of the equation: x2+4x=&#x2212;52" role="presentation">x2+4x=−52x2+4x=−52x^2 + 4x = -\frac{5}{2}52" role="presentation">5252\frac{5}{2} to the right-hand side of the equation:
3. Add and subtract (4/2)2=4" role="presentation">(4/2)2=4(4/2)2=4(4/2)^2 = 4 on the right-hand side of the equation: x2+4x+4&#x2212;4=&#x2212;52" role="presentation">x2+4x+4−4=−52x2+4x+4−4=−52x^2 + 4x + 4 - 4 = -\frac{5}{2}(4/2)2=4" role="presentation" style="position: relative;">(4/2)2=4(4/2)2=4(4/2)^2 = 4 on the right-hand side of the equation: $$x^2 + 4x +
4. Simplify the right-hand side of the equation by factoring the perfect square trinomial: (x+2)2&#x2212;4=&#x2212;52" role="presentation">(x+2)2−4=−52(x+2)2−4=−52(x+2)^2 - 4 = -\frac{5}{2}
5. Take the square root of both sides of the equation and solve for x" role="presentation">xxx: x+2=&#x00B1;32" role="presentation">x+2=±32−−√x+2=±32x+2 = \pm\sqrt{\frac{3}{2}}
Solving for x" role="presentation">xxx, we get: x=&#x2212;2&#x00B1;32" role="presentation">x=−2±32−−√x=−2±32x = -2 \pm \sqrt{\frac{3}{2}}x" role="presentation">xxx: x+2=&#x00B1;32" role="presentation" style="text-align: center; position: relative;">x+2=±32−−√x+2=±32x+2 = \pm\sqrt{\frac{3}{2}}
Solving for x" role="presentation" style="position: relative;">xxx,

So the solutions to the equation 2x2+8x+5=0" role="presentation" style="position: relative;">2x2+8x+5=02x2+8x+5=02x^2+8x+5=0 are x=&#x2212;2+32" role="presentation" style="position: relative;">x=−2+32−−√x=−2+32x=-2+\sqrt{\frac{3}{2}} and x=&#x2212;2&#x2212;32" role="presentation" style="position: relative;">x=−2−32−−√x=−2−32x=-2-\sqrt{\frac{3}{2}}.2x2+8x+5=0" role="presentation" style="position: relative;">2x2+8x+5=02x2+8x+5=02x^2+8x+5=0 are $x=-2+\sqrt{\frac{

I hope this helps! Let me know if you have any other questions.
물론 사각형 완성을 도와드리겠습니다!ax2+bx+c" role="presentation" style="위치: 상대적;">ax2+bx+cax2+bx+cax^2+bx+c, 다음 단계를 따르세요:ax2+bx+c" role="presentation" style="위치: 상대적;">ax2+bx+cax2+bx+cax^2+bx+c, 다음 단계를 따르세요.1. a" role="presentation" style="위치: relative;">aaa (만약 $a\neq방정식의 양변을 a" role="presentation" style="위치: relative;">aaa (만약 a&#x2260;1" role="presentation" style="위치: relative;">a≠1a≠1a\neq 1)로 나누어 제곱 항의 계수를 1로 구합니다.2. 상수 항 c" role="presentation">ccc를 방정식의 오른쪽으로 이동합니다.c" role="presentation" style="위치: 상대적;">ccc3. 방정식의 오른쪽에 (b/2a)2를 더하고 빼기"role="presentation" style="위치: 상대적;">(b/2a)2(b/2a)2(b/2a)^2를 더합니다. 이 단계는 방정식의 오른쪽에 완벽한 정사각형 삼항식을 생성하므로 "정사각형을 완성"하는 단계입니다. (b/2a)2" role="presentation" style="위치: 상대;">(b/2a)2(b/2a)2(b/2a)^2를 방정식의 오른쪽에 표시합니다. 이 단계는 방정식의 오른쪽에 완벽한 정사각형 삼항식을 만들기 때문에 "정사각형을 완성"하는 단계입니다. (b/2a)2" role="presentation">(b/2a)2(b/2a)2(b/2a)^2를 오른쪽에 만듭니다.4. 완전 정사각형 삼항식을 인수분해하여 방정식의 오른쪽을 단순화합니다.5. 방정식 양쪽의 제곱근을 구하여 x"role="presentation" style="위치: 상대;">xxx를 구합니다.다음은 이 과정을 설명하기 위한 예시입니다. 이차식 2x2+8x+5" role="presentation" style="위치: 상대;">2x2+8x+52x2+8x+52x^2 + 8x + 5의 제곱을 완성하고 싶다고 가정합니다. 다음과 같이 진행하면 됩니다:1. 12(2x2+8x+5)=12(0)" role="presentation" style="text-align: 가운데; 위치: 상대;">12(2x2+8x+5)=12(0)12(2x2+8x+5)=12(0)\frac{1}{2}(2x^2+8x+5) = \frac{1}{2}(0)이는 다음과 같이 단순화됩니다: $$x^2 + 4x + \frac{12(2x2+8x+5)=12(0)" role="presentation" style="text-align: center; position: relative;">12(2x2+8x+5)=12(0)12(2x2+8x+5)=12(0)\frac{1}{2}(2x^2+8x+5) = \frac{1}{2}(0)방정식의 양쪽을 2로 나누면 다음과 같이 단순화됩니다: 12(2x2+8x+5)=12(0)" role="presentation">12(2x2+8x+5)=12(0)12(2x2+8x+5)=12(0)\frac{1}{2}(2x^2+8x+5) = \frac{1}{2}(0)이는 다음과 같이 단순화됩니다: x2+4x+52=0" role="presentation">x2+4x+52=0x2+4x+52=0x^2 + 4x + \frac{5}{2} = 02. 상수 항 52" role="presentation">5252\frac{5}{2}를 방정식의 오른쪽으로 이동합니다: x2+4x=&#x2212;52" role="presentation">x2+4x=-52x2+4x=-52x^2 + 4x = -\frac{5}{2}52" role="presentation">5252\frac{5}{2}를 방정식의 오른쪽에 놓습니다:3. 방정식의 오른쪽에 (4/2)2=4"role="presentation">(4/2)2=4(4/2)2=4(4/2)^2 = 4를 더하고 뺍니다: x2+4x+4&#x2212;4=&#x2212;52" role="presentation">x2+4x+4-4=-52x2+4x+4-4=-52x^2 + 4x + 4 - 4 = -\frac{5}{2}(4/2)2=4" role="presentation" style="position: 상대;">(4/2)2=4(4/2)2=4(4/2)^2 = 방정식의 오른쪽에 $$x^2 + 4x + 44. 완전 정사각형 삼항식을 인수분해하여 방정식의 오른쪽을 단순화합니다: (x+2)2&#x2212;4=&#x2212;52" role="presentation">(x+2)2-4=-52(x+2)2-4=-52(x+2)^2 - 4 = -\frac{5}{2}5. 방정식 양쪽의 제곱근을 구하고 x" role="presentation">xxx: x+2=&#x00B1;32" role="presentation">x+2=±32--√x+2=±32x+2 = \pm\sqrt{\frac{3}{2}}x" role="presentation">xxx를 풀면 x=&#x2212;2&#x00B1;32" role="presentation">x=-2±32--√x=-2±32x = -2 \pm \sqrt{\frac{3}{2}}x" role="presentation">xxx를 얻습니다: x+2=&#x00B1;32" role="presentation" style="text-align: 가운데; 위치: 상대적;">x+2=±32--√x+2=±32x+2 = \pm\sqrt{\frac{3}{2}}x에 대한 풀이" role="presentation" style="위치: 상대;">xxx,따라서 방정식 2x2+8x+5=0" role="presentation" style="위치: 상대적;">2x2+8x+5=02x2+8x+5=02x^2+8x+5=0의 해는 x=&#x2212;2+32" role="presentation" style="위치" 입니다: relative;">x=−2+32−−√x=−2+32x=-2+\sqrt{\frac{3}{2}} and x=&#x2212;2&#x2212;32" role="presentation" style="position: relative;">x=−2−32−−√x=−2−32x=-2-\sqrt{\frac{3}{2}}. 2x2+8x+5=0" role="presentation" style="위치: 상대;">2x2+8x+5=02x2+8x+5=02x^2+8x+5=0은 $x=-2+\sqrt{\frac{{2}}입니다.도움이 되었기를 바랍니다! 다른 질문이 있으시면 언제든지 알려주세요.