3n^2-95=4n

Solution for 3n^2-95=4n equation:

3n^2-95=4n

We move all terms to the left:

3n^2-95-(4n)=0

a = 3; b = -4; c = -95;
Δ = b2-4ac
Δ = -42-4·3·(-95)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-34}{2*3}=\frac{-30}{6}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+34}{2*3}=\frac{38}{6}
=6+1/3
$