2m^2+15m-25=0

Solution for 2m^2+15m-25=0 equation:

2m^2+15m-25=0

a = 2; b = 15; c = -25;
Δ = b2-4ac
Δ = 152-4·2·(-25)
Δ = 425
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{17}}{2*2}=\frac{-15-5\sqrt{17}}{4}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{17}}{2*2}=\frac{-15+5\sqrt{17}}{4}
$