r^2+24=11r

Solution for r^2+24=11r equation:

r^2+24=11r

We move all terms to the left:

r^2+24-(11r)=0

a = 1; b = -11; c = +24;
Δ = b2-4ac
Δ = -112-4·1·24
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*1}=\frac{6}{2}
=3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*1}=\frac{16}{2}
=8
$