2x^2+44x-20=0

Solution for 2x^2+44x-20=0 equation:

2x^2+44x-20=0

a = 2; b = 44; c = -20;
Δ = b2-4ac
Δ = 442-4·2·(-20)
Δ = 2096
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:

$\sqrt{\Delta}=\sqrt{2096}=\sqrt{16*131}=\sqrt{16}*\sqrt{131}=4\sqrt{131}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{131}}{2*2}=\frac{-44-4\sqrt{131}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{131}}{2*2}=\frac{-44+4\sqrt{131}}{4}
$