2x^2-32x+58=0

Solution for 2x^2-32x+58=0 equation:

2x^2-32x+58=0

a = 2; b = -32; c = +58;
Δ = b2-4ac
Δ = -322-4·2·58
Δ = 560
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:

$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{35}}{2*2}=\frac{32-4\sqrt{35}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{35}}{2*2}=\frac{32+4\sqrt{35}}{4}
$