40x^2+16x-8=0

Solution for 40x^2+16x-8=0 equation:

40x^2+16x-8=0

a = 40; b = 16; c = -8;
Δ = b2-4ac
Δ = 162-4·40·(-8)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{6}}{2*40}=\frac{-16-16\sqrt{6}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{6}}{2*40}=\frac{-16+16\sqrt{6}}{80}
$