(q)=q^2+4q-12

Solution for (q)=q^2+4q-12 equation:

(q)=q^2+4q-12

We move all terms to the left:

(q)-(q^2+4q-12)=0

We get rid of parentheses

-q^2+q-4q+12=0

We add all the numbers together, and all the variables

-1q^2-3q+12=0

a = -1; b = -3; c = +12;
Δ = b2-4ac
Δ = -32-4·(-1)·12
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{57}}{2*-1}=\frac{3-\sqrt{57}}{-2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*-1}=\frac{3+\sqrt{57}}{-2}
$