3x-4=(4x-3)(4x-3)

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Solution for 3x-4=(4x-3)(4x-3) equation: 



3x-4=(4x-3)(4x-3)

We move all terms to the left:

3x-4-((4x-3)(4x-3))=0

We multiply parentheses ..

-((+16x^2-12x-12x+9))+3x-4=0



We calculate terms in parentheses: -((+16x^2-12x-12x+9)), so:

(+16x^2-12x-12x+9)

We get rid of parentheses

16x^2-12x-12x+9

We add all the numbers together, and all the variables

16x^2-24x+9

Back to the equation:

-(16x^2-24x+9)



We add all the numbers together, and all the variables

3x-(16x^2-24x+9)-4=0

We get rid of parentheses

-16x^2+3x+24x-9-4=0

We add all the numbers together, and all the variables

-16x^2+27x-13=0

a = -16; b = 27; c = -13;
Δ = b2-4ac
Δ = 272-4·(-16)·(-13)
Δ = -103
Delta is less than zero, so there is no solution for the equation

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