3x^2=34x+112

Solution for 3x^2=34x+112 equation:

3x^2=34x+112

We move all terms to the left:

3x^2-(34x+112)=0

We get rid of parentheses

3x^2-34x-112=0

a = 3; b = -34; c = -112;
Δ = b2-4ac
Δ = -342-4·3·(-112)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-50}{2*3}=\frac{-16}{6}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+50}{2*3}=\frac{84}{6}
=14
$