4n^2+26n-65=0

Solution for 4n^2+26n-65=0 equation:

4n^2+26n-65=0

a = 4; b = 26; c = -65;
Δ = b2-4ac
Δ = 262-4·4·(-65)
Δ = 1716
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:

$\sqrt{\Delta}=\sqrt{1716}=\sqrt{4*429}=\sqrt{4}*\sqrt{429}=2\sqrt{429}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{429}}{2*4}=\frac{-26-2\sqrt{429}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{429}}{2*4}=\frac{-26+2\sqrt{429}}{8}
$