Solving for x with exponents (algebra) So I am trying to help a friend do her homework and I am a bit stuck. 
$$8x+3 = 3x^2$$
I can look at this and see that the answer is $3$, but I am having a hard time remembering how to solve for $x$ in this situation. 
Could someone be so kind as to break down the steps in solving for $x$.
Thanks in advance for replies. 

$8x+3=3x^2$

You can solve the above equation using "Splitting middle term of quadratic equation" formula as well. It is mostly useful for simple equations like above.
Solution: 
You can re-write the above as:
$3 \cdot x^2 - 8 \cdot x -3 = 0$
Now try to express the middle term: $8 \cdot x$ as the factor of the product of the coefficient of the other two terms: $3 \cdot (-3) = (-9)$
Now $9$ can be represent as $9 \cdot 1$
Now look at the sign of the middle term: $8 \cdot x$ and it is $-$ (negetive)
Since it is negetive, express $-8x$ as $(-9x + 1x)$
So, the equation in this case will be:
$3 \cdot x^2 -9 \cdot x + 1 \cdot x  -3 = 0$
$\implies (x-3)(3x+1) = 0$
$\implies x =3, -\frac{1}{3}$
Now x can't be negative.
So, $x = 3$
Now put x =3 in the above mentioned equations:
So, 
$lm = 2 \cdot x +1 = 7$

$mn = 6 \cdot x -3 = 15 $ and

$ln = 3 \cdot x^2 -5 = 22$

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Reference: http://www.teacherschoice.com.au/Maths_Library/Algebra/Alg_18.htm
