2x^2+9x-2250=0

Solution for 2x^2+9x-2250=0 equation:

2x^2+9x-2250=0

a = 2; b = 9; c = -2250;
Δ = b2-4ac
Δ = 92-4·2·(-2250)
Δ = 18081
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:

$\sqrt{\Delta}=\sqrt{18081}=\sqrt{441*41}=\sqrt{441}*\sqrt{41}=21\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-21\sqrt{41}}{2*2}=\frac{-9-21\sqrt{41}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+21\sqrt{41}}{2*2}=\frac{-9+21\sqrt{41}}{4}
$