2x^2+23x-1500=0

Solution for 2x^2+23x-1500=0 equation:

2x^2+23x-1500=0

a = 2; b = 23; c = -1500;
Δ = b2-4ac
Δ = 232-4·2·(-1500)
Δ = 12529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{12529}}{2*2}=\frac{-23-\sqrt{12529}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{12529}}{2*2}=\frac{-23+\sqrt{12529}}{4}
$