M(3-4m)=4(8-m^2)

Solution for M(3-4m)=4(8-m^2) equation:

(3-4M)=4(8-M^2)

We move all terms to the left:

(3-4M)-(4(8-M^2))=0

We add all the numbers together, and all the variables

-(4(8-M^2))+(-4M+3)=0

We get rid of parentheses

-(4(8-M^2))-4M+3=0


We calculate terms in parentheses: -(4(8-M^2)), so:

4(8-M^2)

We multiply parentheses

-4M^2+32

Back to the equation:

-(-4M^2+32)


We get rid of parentheses

4M^2-4M-32+3=0

We add all the numbers together, and all the variables

4M^2-4M-29=0

a = 4; b = -4; c = -29;
Δ = b2-4ac
Δ = -42-4·4·(-29)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{30}}{2*4}=\frac{4-4\sqrt{30}}{8}
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{30}}{2*4}=\frac{4+4\sqrt{30}}{8}
$