3m^2+30m=168

Solution for 3m^2+30m=168 equation:

3m^2+30m=168

We move all terms to the left:

3m^2+30m-(168)=0

a = 3; b = 30; c = -168;
Δ = b2-4ac
Δ = 302-4·3·(-168)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-54}{2*3}=\frac{-84}{6}
=-14
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+54}{2*3}=\frac{24}{6}
=4
$