2n^2+n-15=25

Solution for 2n^2+n-15=25 equation:

2n^2+n-15=25

We move all terms to the left:

2n^2+n-15-(25)=0

We add all the numbers together, and all the variables

2n^2+n-40=0

a = 2; b = 1; c = -40;
Δ = b2-4ac
Δ = 12-4·2·(-40)
Δ = 321
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{321}}{2*2}=\frac{-1-\sqrt{321}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{321}}{2*2}=\frac{-1+\sqrt{321}}{4}
$