4x-1=(3x-2)(3x-2)

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

4x-1=(3x-2)(3x-2)

We move all terms to the left:

4x-1-((3x-2)(3x-2))=0

We multiply parentheses ..

-((+9x^2-6x-6x+4))+4x-1=0


We calculate terms in parentheses: -((+9x^2-6x-6x+4)), so:

(+9x^2-6x-6x+4)

We get rid of parentheses

9x^2-6x-6x+4

We add all the numbers together, and all the variables

9x^2-12x+4

Back to the equation:

-(9x^2-12x+4)


We add all the numbers together, and all the variables

4x-(9x^2-12x+4)-1=0

We get rid of parentheses

-9x^2+4x+12x-4-1=0

We add all the numbers together, and all the variables

-9x^2+16x-5=0

a = -9; b = 16; c = -5;
Δ = b2-4ac
Δ = 162-4·(-9)·(-5)
Δ = 76
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{19}}{2*-9}=\frac{-16-2\sqrt{19}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{19}}{2*-9}=\frac{-16+2\sqrt{19}}{-18}
$