-3x^2+340x=0

Solution for -3x^2+340x=0 equation:

-3x^2+340x=0

a = -3; b = 340; c = 0;
Δ = b2-4ac
Δ = 3402-4·(-3)·0
Δ = 115600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{115600}=340$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(340)-340}{2*-3}=\frac{-680}{-6}
=113+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(340)+340}{2*-3}=\frac{0}{-6}
=0
$