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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+6M^2-2M)+4M^2+19)+M=0


We calculate terms in parentheses: -((+6M^2-2M)+4M^2+19), so:

(+6M^2-2M)+4M^2+19

We add all the numbers together, and all the variables

4M^2+(+6M^2-2M)+19

We get rid of parentheses

4M^2+6M^2-2M+19

We add all the numbers together, and all the variables

10M^2-2M+19

Back to the equation:

-(10M^2-2M+19)


We add all the numbers together, and all the variables

M-(10M^2-2M+19)=0

We get rid of parentheses

-10M^2+M+2M-19=0

We add all the numbers together, and all the variables

-10M^2+3M-19=0

a = -10; b = 3; c = -19;
Δ = b2-4ac
Δ = 32-4·(-10)·(-19)
Δ = -751
Delta is less than zero, so there is no solution for the equation