3x^2-36x+95=0

Solution for 3x^2-36x+95=0 equation:

3x^2-36x+95=0

a = 3; b = -36; c = +95;
Δ = b2-4ac
Δ = -362-4·3·95
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-2\sqrt{39}}{2*3}=\frac{36-2\sqrt{39}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+2\sqrt{39}}{2*3}=\frac{36+2\sqrt{39}}{6}
$