r^2+24=10r

Solution for r^2+24=10r equation:

r^2+24=10r

We move all terms to the left:

r^2+24-(10r)=0

a = 1; b = -10; c = +24;
Δ = b2-4ac
Δ = -102-4·1·24
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*1}=\frac{8}{2}
=4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*1}=\frac{12}{2}
=6
$