(4x+1)(4x-1)=9

Solution for (4x+1)(4x-1)=9 equation:

(4x+1)(4x-1)=9

We move all terms to the left:

(4x+1)(4x-1)-(9)=0

We use the square of the difference formula

16x^2-1-9=0

We add all the numbers together, and all the variables

16x^2-10=0

a = 16; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·16·(-10)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{10}}{2*16}=\frac{0-8\sqrt{10}}{32}
=-\frac{8\sqrt{10}}{32}
=-\frac{\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{10}}{2*16}=\frac{0+8\sqrt{10}}{32}
=\frac{8\sqrt{10}}{32}
=\frac{\sqrt{10}}{4}
$