x^2+40=14x

Solution for x^2+40=14x equation:

x^2+40=14x

We move all terms to the left:

x^2+40-(14x)=0

a = 1; b = -14; c = +40;
Δ = b2-4ac
Δ = -142-4·1·40
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6}{2*1}=\frac{8}{2}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6}{2*1}=\frac{20}{2}
=10
$