3x^2=50x+133

Solution for 3x^2=50x+133 equation:

3x^2=50x+133

We move all terms to the left:

3x^2-(50x+133)=0

We get rid of parentheses

3x^2-50x-133=0

a = 3; b = -50; c = -133;
Δ = b2-4ac
Δ = -502-4·3·(-133)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-64}{2*3}=\frac{-14}{6}
=-2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+64}{2*3}=\frac{114}{6}
=19
$