2x^2+25x-25=0

Solution for 2x^2+25x-25=0 equation:

2x^2+25x-25=0

a = 2; b = 25; c = -25;
Δ = b2-4ac
Δ = 252-4·2·(-25)
Δ = 825
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:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{33}}{2*2}=\frac{-25-5\sqrt{33}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{33}}{2*2}=\frac{-25+5\sqrt{33}}{4}
$