32=5x(2x+4)=20

Solution for 32=5x(2x+4)=20 equation:

32=5x(2x+4)=20

We move all terms to the left:

32-(5x(2x+4))=0


We calculate terms in parentheses: -(5x(2x+4)), so:

5x(2x+4)

We multiply parentheses

10x^2+20x

Back to the equation:

-(10x^2+20x)


We get rid of parentheses

-10x^2-20x+32=0

a = -10; b = -20; c = +32;
Δ = b2-4ac
Δ = -202-4·(-10)·32
Δ = 1680
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{105}}{2*-10}=\frac{20-4\sqrt{105}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{105}}{2*-10}=\frac{20+4\sqrt{105}}{-20}
$