20-(M+3)=6(6m-4)m=1

Solution for 20-(M+3)=6(6m-4)m=1 equation:

20-(+3)=6(6M-4)M=1

We move all terms to the left:

20-(+3)-(6(6M-4)M)=0

We add all the numbers together, and all the variables

-(6(6M-4)M)+20-3=0

We add all the numbers together, and all the variables

-(6(6M-4)M)+17=0


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

6(6M-4)M

We multiply parentheses

36M^2-24M

Back to the equation:

-(36M^2-24M)


We get rid of parentheses

-36M^2+24M+17=0

a = -36; b = 24; c = +17;
Δ = b2-4ac
Δ = 242-4·(-36)·17
Δ = 3024
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:

$\sqrt{\Delta}=\sqrt{3024}=\sqrt{144*21}=\sqrt{144}*\sqrt{21}=12\sqrt{21}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-12\sqrt{21}}{2*-36}=\frac{-24-12\sqrt{21}}{-72}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+12\sqrt{21}}{2*-36}=\frac{-24+12\sqrt{21}}{-72}
$