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

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

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

We move all terms to the left:

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

We use the square of the difference formula

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


We calculate terms in parentheses: -((4x-1)2), so:

(4x-1)2

We multiply parentheses

8x-2

Back to the equation:

-(8x-2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

9x^2-8x-14=0

a = 9; b = -8; c = -14;
Δ = b2-4ac
Δ = -82-4·9·(-14)
Δ = 568
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:

$\sqrt{\Delta}=\sqrt{568}=\sqrt{4*142}=\sqrt{4}*\sqrt{142}=2\sqrt{142}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{142}}{2*9}=\frac{8-2\sqrt{142}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{142}}{2*9}=\frac{8+2\sqrt{142}}{18}
$