3v^2+34v=24

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

3v^2+34v=24

We move all terms to the left:

3v^2+34v-(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:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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