3(x2+2x-3)=4(4x2-7x+5)

Solution for 3(x2+2x-3)=4(4x2-7x+5) equation:

3(x2+2x-3)=4(4x^2-7x+5)

We move all terms to the left:

3(x2+2x-3)-(4(4x^2-7x+5))=0

We add all the numbers together, and all the variables

3(+x^2+2x-3)-(4(4x^2-7x+5))=0

We multiply parentheses

3x^2+6x-(4(4x^2-7x+5))-9=0


We calculate terms in parentheses: -(4(4x^2-7x+5)), so:

4(4x^2-7x+5)

We multiply parentheses

16x^2-28x+20

Back to the equation:

-(16x^2-28x+20)


We get rid of parentheses

3x^2-16x^2+6x+28x-20-9=0

We add all the numbers together, and all the variables

-13x^2+34x-29=0

a = -13; b = 34; c = -29;
Δ = b2-4ac
Δ = 342-4·(-13)·(-29)
Δ = -352
Delta is less than zero, so there is no solution for the equation