5m(3m+4)=9+(6m-7)

Solution for 5m(3m+4)=9+(6m-7) equation:

5m(3m+4)=9+(6m-7)

We move all terms to the left:

5m(3m+4)-(9+(6m-7))=0

We multiply parentheses

15m^2+20m-(9+(6m-7))=0


We calculate terms in parentheses: -(9+(6m-7)), so:

9+(6m-7)

determiningTheFunctionDomain
(6m-7)+9

We get rid of parentheses

6m-7+9

We add all the numbers together, and all the variables

6m+2

Back to the equation:

-(6m+2)


We get rid of parentheses

15m^2+20m-6m-2=0

We add all the numbers together, and all the variables

15m^2+14m-2=0

a = 15; b = 14; c = -2;
Δ = b2-4ac
Δ = 142-4·15·(-2)
Δ = 316
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:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{79}}{2*15}=\frac{-14-2\sqrt{79}}{30}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{79}}{2*15}=\frac{-14+2\sqrt{79}}{30}
$