2x^2-16x+13=0

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

2x^2-16x+13=0

a = 2; b = -16; c = +13;
Δ = b2-4ac
Δ = -162-4·2·13
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{-(-16)-2\sqrt{38}}{2*2}=\frac{16-2\sqrt{38}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{38}}{2*2}=\frac{16+2\sqrt{38}}{4}
$