4x^2+8x-126=0

Solution for 4x^2+8x-126=0 equation:

4x^2+8x-126=0

a = 4; b = 8; c = -126;
Δ = b2-4ac
Δ = 82-4·4·(-126)
Δ = 2080
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:

$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{130}}{2*4}=\frac{-8-4\sqrt{130}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{130}}{2*4}=\frac{-8+4\sqrt{130}}{8}
$