5x+20=(x+4)x

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

5x+20=(x+4)x

We move all terms to the left:

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


We calculate terms in parentheses: -((x+4)x), so:

(x+4)x

We multiply parentheses

x^2+4x

Back to the equation:

-(x^2+4x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+x+20=0

a = -1; b = 1; c = +20;
Δ = b2-4ac
Δ = 12-4·(-1)·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*-1}=\frac{-10}{-2}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*-1}=\frac{8}{-2}
=-4
$