10x+20=(2x+4)(x+2)

Solution for 10x+20=(2x+4)(x+2) equation:

10x+20=(2x+4)(x+2)

We move all terms to the left:

10x+20-((2x+4)(x+2))=0

We multiply parentheses ..

-((+2x^2+4x+4x+8))+10x+20=0


We calculate terms in parentheses: -((+2x^2+4x+4x+8)), so:

(+2x^2+4x+4x+8)

We get rid of parentheses

2x^2+4x+4x+8

We add all the numbers together, and all the variables

2x^2+8x+8

Back to the equation:

-(2x^2+8x+8)


We add all the numbers together, and all the variables

10x-(2x^2+8x+8)+20=0

We get rid of parentheses

-2x^2+10x-8x-8+20=0

We add all the numbers together, and all the variables

-2x^2+2x+12=0

a = -2; b = 2; c = +12;
Δ = b2-4ac
Δ = 22-4·(-2)·12
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*-2}=\frac{-12}{-4}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*-2}=\frac{8}{-4}
=-2
$