(x-3)2+6(x+2)=(2x-1)(x+4)+37

Solution for (x-3)2+6(x+2)=(2x-1)(x+4)+37 equation:

(x-3)2+6(x+2)=(2x-1)(x+4)+37

We move all terms to the left:

(x-3)2+6(x+2)-((2x-1)(x+4)+37)=0

We multiply parentheses

2x+6x-((2x-1)(x+4)+37)-6+12=0

We multiply parentheses ..

-((+2x^2+8x-1x-4)+37)+2x+6x-6+12=0


We calculate terms in parentheses: -((+2x^2+8x-1x-4)+37), so:

(+2x^2+8x-1x-4)+37

We get rid of parentheses

2x^2+8x-1x-4+37

We add all the numbers together, and all the variables

2x^2+7x+33

Back to the equation:

-(2x^2+7x+33)


We add all the numbers together, and all the variables

8x-(2x^2+7x+33)+6=0

We get rid of parentheses

-2x^2+8x-7x-33+6=0

We add all the numbers together, and all the variables

-2x^2+x-27=0

a = -2; b = 1; c = -27;
Δ = b2-4ac
Δ = 12-4·(-2)·(-27)
Δ = -215
Delta is less than zero, so there is no solution for the equation