-2x^2+48x+50=0

Solution for -2x^2+48x+50=0 equation:

-2x^2+48x+50=0

a = -2; b = 48; c = +50;
Δ = b2-4ac
Δ = 482-4·(-2)·50
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-52}{2*-2}=\frac{-100}{-4}
=+25
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+52}{2*-2}=\frac{4}{-4}
=-1
$