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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-1x=0

a = 1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·1·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*1}=\frac{2}{2}
=1
$