3x^2+6=31x-50

Solution for 3x^2+6=31x-50 equation:

3x^2+6=31x-50

We move all terms to the left:

3x^2+6-(31x-50)=0

We get rid of parentheses

3x^2-31x+50+6=0

We add all the numbers together, and all the variables

3x^2-31x+56=0

a = 3; b = -31; c = +56;
Δ = b2-4ac
Δ = -312-4·3·56
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-17}{2*3}=\frac{14}{6}
=2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+17}{2*3}=\frac{48}{6}
=8
$