2x^2+10x-2000=0

Solution for 2x^2+10x-2000=0 equation:

2x^2+10x-2000=0

a = 2; b = 10; c = -2000;
Δ = b2-4ac
Δ = 102-4·2·(-2000)
Δ = 16100
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:

$\sqrt{\Delta}=\sqrt{16100}=\sqrt{100*161}=\sqrt{100}*\sqrt{161}=10\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{161}}{2*2}=\frac{-10-10\sqrt{161}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{161}}{2*2}=\frac{-10+10\sqrt{161}}{4}
$