2x^2+2x+8=40

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

2x^2+2x+8=40

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2+2x-32=0

a = 2; b = 2; c = -32;
Δ = b2-4ac
Δ = 22-4·2·(-32)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{65}}{2*2}=\frac{-2-2\sqrt{65}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{65}}{2*2}=\frac{-2+2\sqrt{65}}{4}
$