x^2+6x-95=0

Solution for x^2+6x-95=0 equation:

x^2+6x-95=0

a = 1; b = 6; c = -95;
Δ = b2-4ac
Δ = 62-4·1·(-95)
Δ = 416
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:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{26}}{2*1}=\frac{-6-4\sqrt{26}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{26}}{2*1}=\frac{-6+4\sqrt{26}}{2}
$