n^2+14=-10n

Solution for n^2+14=-10n equation:

n^2+14=-10n

We move all terms to the left:

n^2+14-(-10n)=0

We get rid of parentheses

n^2+10n+14=0

a = 1; b = 10; c = +14;
Δ = b2-4ac
Δ = 102-4·1·14
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{11}}{2*1}=\frac{-10-2\sqrt{11}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{11}}{2*1}=\frac{-10+2\sqrt{11}}{2}
$