40=2x^2-8x-5

Solution for 40=2x^2-8x-5 equation:

40=2x^2-8x-5

We move all terms to the left:

40-(2x^2-8x-5)=0

We get rid of parentheses

-2x^2+8x+5+40=0

We add all the numbers together, and all the variables

-2x^2+8x+45=0

a = -2; b = 8; c = +45;
Δ = b2-4ac
Δ = 82-4·(-2)·45
Δ = 424
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:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{106}}{2*-2}=\frac{-8-2\sqrt{106}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{106}}{2*-2}=\frac{-8+2\sqrt{106}}{-4}
$