(x^2+3x-7)=(8x^2+8x-3)

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

(x^2+3x-7)=(8x^2+8x-3)

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((8x^2+8x-3)), so:

(8x^2+8x-3)

We get rid of parentheses

8x^2+8x-3

Back to the equation:

-(8x^2+8x-3)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-7x^2-5x-4=0

a = -7; b = -5; c = -4;
Δ = b2-4ac
Δ = -52-4·(-7)·(-4)
Δ = -87
Delta is less than zero, so there is no solution for the equation