4x^2+40x-12=0

Solution for 4x^2+40x-12=0 equation:

4x^2+40x-12=0

a = 4; b = 40; c = -12;
Δ = b2-4ac
Δ = 402-4·4·(-12)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-16\sqrt{7}}{2*4}=\frac{-40-16\sqrt{7}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16\sqrt{7}}{2*4}=\frac{-40+16\sqrt{7}}{8}
$