2n^2+25n-13=0

Solution for 2n^2+25n-13=0 equation:

2n^2+25n-13=0

a = 2; b = 25; c = -13;
Δ = b2-4ac
Δ = 252-4·2·(-13)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-27}{2*2}=\frac{-52}{4}
=-13
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+27}{2*2}=\frac{2}{4}
=1/2
$