27-(m+4)=3(6m-4);m=2

Solution for 27-(m+4)=3(6m-4);m=2 equation:

27-(m+4)=3(6m-4)m=2

We move all terms to the left:

27-(m+4)-(3(6m-4)m)=0

We get rid of parentheses

-m-(3(6m-4)m)-4+27=0


We calculate terms in parentheses: -(3(6m-4)m), so:

3(6m-4)m

We multiply parentheses

18m^2-12m

Back to the equation:

-(18m^2-12m)


We add all the numbers together, and all the variables

-1m-(18m^2-12m)+23=0

We get rid of parentheses

-18m^2-1m+12m+23=0

We add all the numbers together, and all the variables

-18m^2+11m+23=0

a = -18; b = 11; c = +23;
Δ = b2-4ac
Δ = 112-4·(-18)·23
Δ = 1777
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{1777}}{2*-18}=\frac{-11-\sqrt{1777}}{-36}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{1777}}{2*-18}=\frac{-11+\sqrt{1777}}{-36}
$