2x+8=4x2+20x+16

Solution for 2x+8=4x2+20x+16 equation:

2x+8=4x^2+20x+16

We move all terms to the left:

2x+8-(4x^2+20x+16)=0

We get rid of parentheses

-4x^2+2x-20x-16+8=0

We add all the numbers together, and all the variables

-4x^2-18x-8=0

a = -4; b = -18; c = -8;
Δ = b2-4ac
Δ = -182-4·(-4)·(-8)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-14}{2*-4}=\frac{4}{-8}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+14}{2*-4}=\frac{32}{-8}
=-4
$