4x^2+8x-92=0

Solution for 4x^2+8x-92=0 equation:

4x^2+8x-92=0

a = 4; b = 8; c = -92;
Δ = b2-4ac
Δ = 82-4·4·(-92)
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{-(8)-16\sqrt{6}}{2*4}=\frac{-8-16\sqrt{6}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16\sqrt{6}}{2*4}=\frac{-8+16\sqrt{6}}{8}
$