8x(2x+4)=20

Solution for 8x(2x+4)=20 equation:

8x(2x+4)=20

We move all terms to the left:

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

We multiply parentheses

16x^2+32x-20=0

a = 16; b = 32; c = -20;
Δ = b2-4ac
Δ = 322-4·16·(-20)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-48}{2*16}=\frac{-80}{32}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+48}{2*16}=\frac{16}{32}
=1/2
$