k^2=7k+30

Solution for k^2=7k+30 equation:

k^2=7k+30

We move all terms to the left:

k^2-(7k+30)=0

We get rid of parentheses

k^2-7k-30=0

a = 1; b = -7; c = -30;
Δ = b2-4ac
Δ = -72-4·1·(-30)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13}{2*1}=\frac{-6}{2}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*1}=\frac{20}{2}
=10
$