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

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

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

We move all terms to the left:

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

We use the square of the difference formula

9x^2-((4x-1))-16=0


We calculate terms in parentheses: -((4x-1)), so:

(4x-1)

We get rid of parentheses

4x-1

Back to the equation:

-(4x-1)


We get rid of parentheses

9x^2-4x+1-16=0

We add all the numbers together, and all the variables

9x^2-4x-15=0

a = 9; b = -4; c = -15;
Δ = b2-4ac
Δ = -42-4·9·(-15)
Δ = 556
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:

$\sqrt{\Delta}=\sqrt{556}=\sqrt{4*139}=\sqrt{4}*\sqrt{139}=2\sqrt{139}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{139}}{2*9}=\frac{4-2\sqrt{139}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{139}}{2*9}=\frac{4+2\sqrt{139}}{18}
$