(x-5)(x+4)=20

Solution for (x-5)(x+4)=20 equation:

(x-5)(x+4)=20

We move all terms to the left:

(x-5)(x+4)-(20)=0

We multiply parentheses ..

(+x^2+4x-5x-20)-20=0

We get rid of parentheses

x^2+4x-5x-20-20=0

We add all the numbers together, and all the variables

x^2-1x-40=0

a = 1; b = -1; c = -40;
Δ = b2-4ac
Δ = -12-4·1·(-40)
Δ = 161
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{161}}{2*1}=\frac{1-\sqrt{161}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{161}}{2*1}=\frac{1+\sqrt{161}}{2}
$