-16x^2+25x+1200=0

Solution for -16x^2+25x+1200=0 equation:

-16x^2+25x+1200=0

a = -16; b = 25; c = +1200;
Δ = b2-4ac
Δ = 252-4·(-16)·1200
Δ = 77425
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:

$\sqrt{\Delta}=\sqrt{77425}=\sqrt{25*3097}=\sqrt{25}*\sqrt{3097}=5\sqrt{3097}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{3097}}{2*-16}=\frac{-25-5\sqrt{3097}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{3097}}{2*-16}=\frac{-25+5\sqrt{3097}}{-32}
$