M(x)=(3x-1)(3x-1)+4x^2+19

Solution for M(x)=(3x-1)(3x-1)+4x^2+19 equation:

(M)=(3M-1)(3M-1)+4M^2+19

We move all terms to the left:

(M)-((3M-1)(3M-1)+4M^2+19)=0

We multiply parentheses ..

-((+9M^2-3M-3M+1)+4M^2+19)+M=0


We calculate terms in parentheses: -((+9M^2-3M-3M+1)+4M^2+19), so:

(+9M^2-3M-3M+1)+4M^2+19

We add all the numbers together, and all the variables

4M^2+(+9M^2-3M-3M+1)+19

We get rid of parentheses

4M^2+9M^2-3M-3M+1+19

We add all the numbers together, and all the variables

13M^2-6M+20

Back to the equation:

-(13M^2-6M+20)


We add all the numbers together, and all the variables

M-(13M^2-6M+20)=0

We get rid of parentheses

-13M^2+M+6M-20=0

We add all the numbers together, and all the variables

-13M^2+7M-20=0

a = -13; b = 7; c = -20;
Δ = b2-4ac
Δ = 72-4·(-13)·(-20)
Δ = -991
Delta is less than zero, so there is no solution for the equation