k^2+21k=72

Solution for k^2+21k=72 equation:

k^2+21k=72

We move all terms to the left:

k^2+21k-(72)=0

a = 1; b = 21; c = -72;
Δ = b2-4ac
Δ = 212-4·1·(-72)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-27}{2*1}=\frac{-48}{2}
=-24
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+27}{2*1}=\frac{6}{2}
=3
$