# How do you solve x^2-x=12?

Feb 3, 2015

We can use the Sum-Product method.

Bring everything to one side:
${x}^{2} - x = 12 \to {x}^{2} - x - 12 = 0$

We now have a quadratic equation of the form
$a {x}^{2} + b x + c = 0$ where $a = 1 , b = - 1 \mathmr{and} c = - 12$

We find two numbers that will give $c = - 12$ as a product and $b = - 1$ as a sum (or difference).
We can try $1 \cdot 12 , 2 \cdot 6 , 3 \cdot 4$

$3 \mathmr{and} 4$ will fit, with a $-$sign to the $4$, as that would make the sum $3 - 4 = - 1$ and the product $3 \cdot - 4 = - 12$

Now we can rewrite the equation:
$\left(x + 3\right) \left(x - 4\right) = 0$

Which leaves us with two possibilities:
$\left(x + 3\right) = 0 \to x = - 3$ OR $\left(x - 4\right) = 0 \to x = 4$

$x = - 3 \mathmr{and} x = 4$