2x^2+5x-200=0

Solution for 2x^2+5x-200=0 equation:

2x^2+5x-200=0

a = 2; b = 5; c = -200;
Δ = b2-4ac
Δ = 52-4·2·(-200)
Δ = 1625
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:

$\sqrt{\Delta}=\sqrt{1625}=\sqrt{25*65}=\sqrt{25}*\sqrt{65}=5\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{65}}{2*2}=\frac{-5-5\sqrt{65}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{65}}{2*2}=\frac{-5+5\sqrt{65}}{4}
$