(x-3)(x+4)=20

Solution for (x-3)(x+4)=20 equation:

(x-3)(x+4)=20

We move all terms to the left:

(x-3)(x+4)-(20)=0

We multiply parentheses ..

(+x^2+4x-3x-12)-20=0

We get rid of parentheses

x^2+4x-3x-12-20=0

We add all the numbers together, and all the variables

x^2+x-32=0

a = 1; b = 1; c = -32;
Δ = b2-4ac
Δ = 12-4·1·(-32)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{129}}{2*1}=\frac{-1-\sqrt{129}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{129}}{2*1}=\frac{-1+\sqrt{129}}{2}
$