4x2+56x-540=0

Solution for 4x2+56x-540=0 equation:

4x^2+56x-540=0

a = 4; b = 56; c = -540;
Δ = b2-4ac
Δ = 562-4·4·(-540)
Δ = 11776
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:

$\sqrt{\Delta}=\sqrt{11776}=\sqrt{256*46}=\sqrt{256}*\sqrt{46}=16\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-16\sqrt{46}}{2*4}=\frac{-56-16\sqrt{46}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+16\sqrt{46}}{2*4}=\frac{-56+16\sqrt{46}}{8}
$