x^2=10x+39

Solution for x^2=10x+39 equation:

x^2=10x+39

We move all terms to the left:

x^2-(10x+39)=0

We get rid of parentheses

x^2-10x-39=0

a = 1; b = -10; c = -39;
Δ = b2-4ac
Δ = -102-4·1·(-39)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-16}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+16}{2*1}=\frac{26}{2}
=13
$