M=(3x-2)(4x-5)=

Solution for M=(3x-2)(4x-5)= equation:

=(3M-2)(4M-5)=

We move all terms to the left:

-((3M-2)(4M-5))=0

We multiply parentheses ..

-((+12M^2-15M-8M+10))=0


We calculate terms in parentheses: -((+12M^2-15M-8M+10)), so:

(+12M^2-15M-8M+10)

We get rid of parentheses

12M^2-15M-8M+10

We add all the numbers together, and all the variables

12M^2-23M+10

Back to the equation:

-(12M^2-23M+10)


We get rid of parentheses

-12M^2+23M-10=0

a = -12; b = 23; c = -10;
Δ = b2-4ac
Δ = 232-4·(-12)·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$M_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*-12}=\frac{-30}{-24}
=1+1/4
$
$M_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*-12}=\frac{-16}{-24}
=2/3
$