-11x^2+490x+3000=0

Solution for -11x^2+490x+3000=0 equation:

-11x^2+490x+3000=0

a = -11; b = 490; c = +3000;
Δ = b2-4ac
Δ = 4902-4·(-11)·3000
Δ = 372100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{372100}=610$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(490)-610}{2*-11}=\frac{-1100}{-22}
=+50
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(490)+610}{2*-11}=\frac{120}{-22}
=-5+5/11
$