-49x^2+50x+90=0

Solution for -49x^2+50x+90=0 equation:

-49x^2+50x+90=0

a = -49; b = 50; c = +90;
Δ = b2-4ac
Δ = 502-4·(-49)·90
Δ = 20140
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:

$\sqrt{\Delta}=\sqrt{20140}=\sqrt{4*5035}=\sqrt{4}*\sqrt{5035}=2\sqrt{5035}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{5035}}{2*-49}=\frac{-50-2\sqrt{5035}}{-98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{5035}}{2*-49}=\frac{-50+2\sqrt{5035}}{-98}
$