x^2=10x+13=4

Solution for x^2=10x+13=4 equation:

x^2=10x+13=4

We move all terms to the left:

x^2-(10x+13)=0

We get rid of parentheses

x^2-10x-13=0

a = 1; b = -10; c = -13;
Δ = b2-4ac
Δ = -102-4·1·(-13)
Δ = 152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{38}}{2*1}=\frac{10-2\sqrt{38}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{38}}{2*1}=\frac{10+2\sqrt{38}}{2}
$