r^2-10r+13=0

Solution for r^2-10r+13=0 equation:

r^2-10r+13=0

a = 1; b = -10; c = +13;
Δ = b2-4ac
Δ = -102-4·1·13
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{3}}{2*1}=\frac{10-4\sqrt{3}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{3}}{2*1}=\frac{10+4\sqrt{3}}{2}
$