r^2+10=-12r

Solution for r^2+10=-12r equation:

r^2+10=-12r

We move all terms to the left:

r^2+10-(-12r)=0

We get rid of parentheses

r^2+12r+10=0

a = 1; b = 12; c = +10;
Δ = b2-4ac
Δ = 122-4·1·10
Δ = 104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{26}}{2*1}=\frac{-12-2\sqrt{26}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{26}}{2*1}=\frac{-12+2\sqrt{26}}{2}
$