-50x^2+35x+99=0

Solution for -50x^2+35x+99=0 equation:

-50x^2+35x+99=0

a = -50; b = 35; c = +99;
Δ = b2-4ac
Δ = 352-4·(-50)·99
Δ = 21025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{21025}=145$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-145}{2*-50}=\frac{-180}{-100}
=1+4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+145}{2*-50}=\frac{110}{-100}
=-1+1/10
$