(2m+1)(3m-2=)

Solution for (2m+1)(3m-2=) equation:

(2m+1)(3m-2=)

We move all terms to the left:

(2m+1)(3m-2-())=0


We calculate terms in parentheses: +(2m+1)(3m-2-()), so:

2m+1)(3m-2-()

We add all the numbers together, and all the variables

2m+1)(3m

Back to the equation:

+(2m+1)(3m)


We multiply parentheses

6m^2+3m=0

a = 6; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·6·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*6}=\frac{-6}{12}
=-1/2
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*6}=\frac{0}{12}
=0
$