In a quadratic equation, one or more variables is squared (x2{\displaystyle x^{2}} or y2{\displaystyle y^{2}}), and there are no higher powers. The lines these equations represent are curved, so they can intersect a straight line at 0, 1, or 2 points. This section will teach you how to find the 0, 1, or 2 solutions to your problem.

Expand equations with parentheses to check whether they're quadratics. For example, y=(x+3)(x){\displaystyle y=(x+3)(x)} is quadratic, since it expands into y=x2+3x.{\displaystyle y=x^{2}+3x.}

Equations for a circle or ellipse have both an x2{\displaystyle x^{2}} and a y2{\displaystyle y^{2}} term.If you're having trouble with these special cases, see the Tips section below.
 If necessary, rewrite each equation so y is alone on one side.


Example: Find the intersection of x2+2x−y=−1{\displaystyle x^{2}+2x-y=-1} and y=x+7{\displaystyle y=x+7}.

Rewrite the quadratic equation in terms of y:


y=x2+2x+1{\displaystyle y=x^{2}+2x+1} and y=x+7{\displaystyle y=x+7}.

This example has one quadratic equation and one linear equation. Problems with two quadratic equations are solved in a similar way.
 Once you've set both equations equal to y, you know the two sides without a y are equal to each other.


Example: y=x2+2x+1{\displaystyle y=x^{2}+2x+1} and y=x+7{\displaystyle y=x+7}


x2+2x+1=x+7{\displaystyle x^{2}+2x+1=x+7}


, Use standard algebraic techniques to get all the terms on one side. This will set the problem up so we can solve it in the next step.


Example: x2+2x+1=x+7{\displaystyle x^{2}+2x+1=x+7}

Subtract x from each side:


x2+x+1=7{\displaystyle x^{2}+x+1=7}

Subtract 7 from each side:


x2+x−6=0{\displaystyle x^{2}+x-6=0}


, Once you've set one side equal to zero, there are three ways to solve a quadratic equation. Different people find different methods easier. You can read about the quadratic formula or "completing the square", or follow along with this example of the factoring method:


Example: x2+x−6=0{\displaystyle x^{2}+x-6=0}

The goal of factoring is to find the two factors that multiply together to make this equation. Starting with the first term, we know x2{\displaystyle x^{2}} can divide into x, and x. Write down (x    )(x    ) = 0 to show this.

The last term is -6. List each pair of factors that multiply to make negative six: −6∗1{\displaystyle -6*1}, −3∗2{\displaystyle -3*2}, −2∗3{\displaystyle -2*3}, and −1∗6{\displaystyle -1*6}.

The middle term is x (which you could write as 1x). Add each pair of factors together until you get 1 as an answer. The correct pair of factors is −2∗3{\displaystyle -2*3}, since −2+3=1{\displaystyle -2+3=1}.

Fill out the gaps in your answer with this pair of factors: (x−2)(x+3)=0{\displaystyle (x-2)(x+3)=0}.
 If you work too quickly, you might find one solution to the problem and not realize there's a second one. Here's how to find the two x-values for lines that intersect at two points:


Example (factoring): We ended up with the equation (x−2)(x+3)=0{\displaystyle (x-2)(x+3)=0}. If either of the factors in parentheses equal 0, the equation is true. One solution is x−2=0{\displaystyle x-2=0} → x=2{\displaystyle x=2}. The other solution is x+3=0{\displaystyle x+3=0} → x=−3{\displaystyle x=-3}.


Example (quadratic equation or complete the square): If you used one of these methods to solve your equation, a square root will show up. For example, our equation becomes x=(−1+25)/2{\displaystyle x=(-1+{\sqrt {25}})/2}. Remember that a square root can simplify to two different solutions: 25=5∗5{\displaystyle {\sqrt {25}}=5*5}, and 25=(−5)∗(−5){\displaystyle {\sqrt {25}}=(-5)*(-5)}. Write two equations, one for each possibility, and solve for x in each one.
 Two lines that barely touch only have one intersection, and two lines that never touch have zero. Here's how to recognize these:

One solution: The problems factor into two identical factors ((x-1)(x-1) = 0). When plugged into the quadratic formula, the square root term is 0{\displaystyle {\sqrt {0}}}. You only need to solve one equation.

No real solution: There are no factors that satisfy the requirements (summing to the middle term). When plugged into the quadratic formula, you get a negative number under the square root sign (such as −2{\displaystyle {\sqrt {-2}}}). Write "no solution" as your answer.
 Once you have the x-value of your intersection, plug it back into one of the equations you started with. Solve for y to find the y-value. If you have a second x-value, repeat for this as well.


Example: We found two solutions, x=2{\displaystyle x=2} and x=−3{\displaystyle x=-3}. One of our lines has the equation y=x+7{\displaystyle y=x+7}. Plug in y=2+7{\displaystyle y=2+7} and y=−3+7{\displaystyle y=-3+7}, then solve each equation to find that y=9{\displaystyle y=9} and y=4{\displaystyle y=4}.
 Now write your answer in coordinate form, with the x-value and y-value of the intersection points. If you have two answers, make sure you match the correct x-value to each y-value.


Example: When we plugged in x=2{\displaystyle x=2}, we got y=9{\displaystyle y=9}, so one intersection is at (2, 9). The same process for our second solution tells us another intersection lies at (-3, 4).
Recognize quadratic equations.
Write the equations in terms of y.
Combine the two equations to cancel out the y.
Arrange the new equation so one side is equal to zero.
Solve the quadratic equation.
Keep an eye out for two solutions for x.
Solve problems with one or zero solutions.
Plug your x-values back into either original equation.
Write the point coordinates.
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