40=20x^2+2x+5

Solution for 40=20x^2+2x+5 equation:

40=20x^2+2x+5

We move all terms to the left:

40-(20x^2+2x+5)=0

We get rid of parentheses

-20x^2-2x-5+40=0

We add all the numbers together, and all the variables

-20x^2-2x+35=0

a = -20; b = -2; c = +35;
Δ = b2-4ac
Δ = -22-4·(-20)·35
Δ = 2804
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:

$\sqrt{\Delta}=\sqrt{2804}=\sqrt{4*701}=\sqrt{4}*\sqrt{701}=2\sqrt{701}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{701}}{2*-20}=\frac{2-2\sqrt{701}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{701}}{2*-20}=\frac{2+2\sqrt{701}}{-40}
$