2x^2+8x+16=40

Solution for 2x^2+8x+16=40 equation:

2x^2+8x+16=40

We move all terms to the left:

2x^2+8x+16-(40)=0

We add all the numbers together, and all the variables

2x^2+8x-24=0

a = 2; b = 8; c = -24;
Δ = b2-4ac
Δ = 82-4·2·(-24)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*2}=\frac{8}{4}
=2
$