a^2-16a+13=0

Solution for a^2-16a+13=0 equation:

a^2-16a+13=0

a = 1; b = -16; c = +13;
Δ = b2-4ac
Δ = -162-4·1·13
Δ = 204
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{51}}{2*1}=\frac{16-2\sqrt{51}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{51}}{2*1}=\frac{16+2\sqrt{51}}{2}
$