(3x^2+7x+12)=(8x^2+x+29)

Solution for (3x^2+7x+12)=(8x^2+x+29) equation:

(3x^2+7x+12)=(8x^2+x+29)

We move all terms to the left:

(3x^2+7x+12)-((8x^2+x+29))=0

We get rid of parentheses

3x^2+7x-((8x^2+x+29))+12=0


We calculate terms in parentheses: -((8x^2+x+29)), so:

(8x^2+x+29)

We get rid of parentheses

8x^2+x+29

Back to the equation:

-(8x^2+x+29)


We get rid of parentheses

3x^2-8x^2+7x-x-29+12=0

We add all the numbers together, and all the variables

-5x^2+6x-17=0

a = -5; b = 6; c = -17;
Δ = b2-4ac
Δ = 62-4·(-5)·(-17)
Δ = -304
Delta is less than zero, so there is no solution for the equation