3x^2+50x-600=0

Solution for 3x^2+50x-600=0 equation:

3x^2+50x-600=0

a = 3; b = 50; c = -600;
Δ = b2-4ac
Δ = 502-4·3·(-600)
Δ = 9700
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:

$\sqrt{\Delta}=\sqrt{9700}=\sqrt{100*97}=\sqrt{100}*\sqrt{97}=10\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{97}}{2*3}=\frac{-50-10\sqrt{97}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{97}}{2*3}=\frac{-50+10\sqrt{97}}{6}
$