n^2+n-131.25=0

Solution for n^2+n-131.25=0 equation:

n^2+n-131.25=0

a = 1; b = 1; c = -131.25;
Δ = b2-4ac
Δ = 12-4·1·(-131.25)
Δ = 526
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{526}}{2*1}=\frac{-1-\sqrt{526}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{526}}{2*1}=\frac{-1+\sqrt{526}}{2}
$