7m+4m(m+3)=5m+6(2m-1)

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Solution for 7m+4m(m+3)=5m+6(2m-1) equation: 



7m+4m(m+3)=5m+6(2m-1)

We move all terms to the left:

7m+4m(m+3)-(5m+6(2m-1))=0

We multiply parentheses

4m^2+7m+12m-(5m+6(2m-1))=0



We calculate terms in parentheses: -(5m+6(2m-1)), so:

5m+6(2m-1)

We multiply parentheses

5m+12m-6

We add all the numbers together, and all the variables

17m-6

Back to the equation:

-(17m-6)



We add all the numbers together, and all the variables

4m^2+19m-(17m-6)=0

We get rid of parentheses

4m^2+19m-17m+6=0

We add all the numbers together, and all the variables

4m^2+2m+6=0

a = 4; b = 2; c = +6;
Δ = b2-4ac
Δ = 22-4·4·6
Δ = -92
Delta is less than zero, so there is no solution for the equation

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