(x+4)(x+1)=5x+20

Solution for (x+4)(x+1)=5x+20 equation:

(x+4)(x+1)=5x+20

We move all terms to the left:

(x+4)(x+1)-(5x+20)=0

We get rid of parentheses

(x+4)(x+1)-5x-20=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-16=0

a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2}
=4
$