n^2+14n=9

Solution for n^2+14n=9 equation:

n^2+14n=9

We move all terms to the left:

n^2+14n-(9)=0

a = 1; b = 14; c = -9;
Δ = b2-4ac
Δ = 142-4·1·(-9)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{58}}{2*1}=\frac{-14-2\sqrt{58}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{58}}{2*1}=\frac{-14+2\sqrt{58}}{2}
$