k^2+10=7k

Solution for k^2+10=7k equation:

k^2+10=7k

We move all terms to the left:

k^2+10-(7k)=0

a = 1; b = -7; c = +10;
Δ = b2-4ac
Δ = -72-4·1·10
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*1}=\frac{4}{2}
=2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*1}=\frac{10}{2}
=5
$