3n^2+72n-70=0

Solution for 3n^2+72n-70=0 equation:

3n^2+72n-70=0

a = 3; b = 72; c = -70;
Δ = b2-4ac
Δ = 722-4·3·(-70)
Δ = 6024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6024}=\sqrt{4*1506}=\sqrt{4}*\sqrt{1506}=2\sqrt{1506}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-2\sqrt{1506}}{2*3}=\frac{-72-2\sqrt{1506}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+2\sqrt{1506}}{2*3}=\frac{-72+2\sqrt{1506}}{6}
$