2n^2+n-78=0

Solution for 2n^2+n-78=0 equation:

2n^2+n-78=0

a = 2; b = 1; c = -78;
Δ = b2-4ac
Δ = 12-4·2·(-78)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-25}{2*2}=\frac{-26}{4}
=-6+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+25}{2*2}=\frac{24}{4}
=6
$