((m^2)/3)-2m=7

Solution for ((m^2)/3)-2m=7 equation:

((m^2)/3)-2m=7

We move all terms to the left:

((m^2)/3)-2m-(7)=0

We add all the numbers together, and all the variables

-2m+(m^2/3)-7=0

We get rid of parentheses

m^2/3-2m-7=0

We multiply all the terms by the denominator


m^2-2m*3-7*3=0

We add all the numbers together, and all the variables

m^2-2m*3-21=0

Wy multiply elements

m^2-6m-21=0

a = 1; b = -6; c = -21;
Δ = b2-4ac
Δ = -62-4·1·(-21)
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{30}}{2*1}=\frac{6-2\sqrt{30}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{30}}{2*1}=\frac{6+2\sqrt{30}}{2}
$