2x^2+32x+26=0

Solution for 2x^2+32x+26=0 equation:

2x^2+32x+26=0

a = 2; b = 32; c = +26;
Δ = b2-4ac
Δ = 322-4·2·26
Δ = 816
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:

$\sqrt{\Delta}=\sqrt{816}=\sqrt{16*51}=\sqrt{16}*\sqrt{51}=4\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{51}}{2*2}=\frac{-32-4\sqrt{51}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{51}}{2*2}=\frac{-32+4\sqrt{51}}{4}
$