a^2+40=13a

Solution for a^2+40=13a equation:

a^2+40=13a

We move all terms to the left:

a^2+40-(13a)=0

a = 1; b = -13; c = +40;
Δ = b2-4ac
Δ = -132-4·1·40
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*1}=\frac{10}{2}
=5
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*1}=\frac{16}{2}
=8
$