4x^2-8x=40

Solution for 4x^2-8x=40 equation:

4x^2-8x=40

We move all terms to the left:

4x^2-8x-(40)=0

a = 4; b = -8; c = -40;
Δ = b2-4ac
Δ = -82-4·4·(-40)
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{11}}{2*4}=\frac{8-8\sqrt{11}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{11}}{2*4}=\frac{8+8\sqrt{11}}{8}
$