24=3x^2+34x

Solution for 24=3x^2+34x equation:

24=3x^2+34x

We move all terms to the left:

24-(3x^2+34x)=0

We get rid of parentheses

-3x^2-34x+24=0

a = -3; b = -34; c = +24;
Δ = b2-4ac
Δ = -342-4·(-3)·24
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-38}{2*-3}=\frac{-4}{-6}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+38}{2*-3}=\frac{72}{-6}
=-12
$