4x^2+10x-95=0

Solution for 4x^2+10x-95=0 equation:

4x^2+10x-95=0

a = 4; b = 10; c = -95;
Δ = b2-4ac
Δ = 102-4·4·(-95)
Δ = 1620
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:

$\sqrt{\Delta}=\sqrt{1620}=\sqrt{324*5}=\sqrt{324}*\sqrt{5}=18\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-18\sqrt{5}}{2*4}=\frac{-10-18\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+18\sqrt{5}}{2*4}=\frac{-10+18\sqrt{5}}{8}
$