4x^2+2x=40

Solution for 4x^2+2x=40 equation:

4x^2+2x=40

We move all terms to the left:

4x^2+2x-(40)=0

a = 4; b = 2; c = -40;
Δ = b2-4ac
Δ = 22-4·4·(-40)
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{161}}{2*4}=\frac{-2-2\sqrt{161}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{161}}{2*4}=\frac{-2+2\sqrt{161}}{8}
$