2x^2+25x-3725=0

Solution for 2x^2+25x-3725=0 equation:

2x^2+25x-3725=0

a = 2; b = 25; c = -3725;
Δ = b2-4ac
Δ = 252-4·2·(-3725)
Δ = 30425
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:

$\sqrt{\Delta}=\sqrt{30425}=\sqrt{25*1217}=\sqrt{25}*\sqrt{1217}=5\sqrt{1217}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{1217}}{2*2}=\frac{-25-5\sqrt{1217}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{1217}}{2*2}=\frac{-25+5\sqrt{1217}}{4}
$