2x^2+24x-728=0

Solution for 2x^2+24x-728=0 equation:

2x^2+24x-728=0

a = 2; b = 24; c = -728;
Δ = b2-4ac
Δ = 242-4·2·(-728)
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-80}{2*2}=\frac{-104}{4}
=-26
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+80}{2*2}=\frac{56}{4}
=14
$