(x+3)(x-4)=120

Solution for (x+3)(x-4)=120 equation:

(x+3)(x-4)=120

We move all terms to the left:

(x+3)(x-4)-(120)=0

We multiply parentheses ..

(+x^2-4x+3x-12)-120=0

We get rid of parentheses

x^2-4x+3x-12-120=0

We add all the numbers together, and all the variables

x^2-1x-132=0

a = 1; b = -1; c = -132;
Δ = b2-4ac
Δ = -12-4·1·(-132)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-23}{2*1}=\frac{-22}{2}
=-11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+23}{2*1}=\frac{24}{2}
=12
$