4x^2+16x-125=0

Solution for 4x^2+16x-125=0 equation:

4x^2+16x-125=0

a = 4; b = 16; c = -125;
Δ = b2-4ac
Δ = 162-4·4·(-125)
Δ = 2256
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:

$\sqrt{\Delta}=\sqrt{2256}=\sqrt{16*141}=\sqrt{16}*\sqrt{141}=4\sqrt{141}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{141}}{2*4}=\frac{-16-4\sqrt{141}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{141}}{2*4}=\frac{-16+4\sqrt{141}}{8}
$