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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((x-4)(x-4)), so:

(x-4)(x-4)

We multiply parentheses ..

(+x^2-4x-4x+16)

We get rid of parentheses

x^2-4x-4x+16

We add all the numbers together, and all the variables

x^2-8x+16

Back to the equation:

-(x^2-8x+16)


We get rid of parentheses

16x^2-x^2-12x-12x+8x+9-16=0

We add all the numbers together, and all the variables

15x^2-16x-7=0

a = 15; b = -16; c = -7;
Δ = b2-4ac
Δ = -162-4·15·(-7)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-26}{2*15}=\frac{-10}{30}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+26}{2*15}=\frac{42}{30}
=1+2/5
$