2x^2+8x-73=0

Solution for 2x^2+8x-73=0 equation:

2x^2+8x-73=0

a = 2; b = 8; c = -73;
Δ = b2-4ac
Δ = 82-4·2·(-73)
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-18\sqrt{2}}{2*2}=\frac{-8-18\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+18\sqrt{2}}{2*2}=\frac{-8+18\sqrt{2}}{4}
$