64x^2=(2-3x)(2+3x)

Solution for 64x^2=(2-3x)(2+3x) equation:

64x^2=(2-3x)(2+3x)

We move all terms to the left:

64x^2-((2-3x)(2+3x))=0

We add all the numbers together, and all the variables

64x^2-((-3x+2)(3x+2))=0

We multiply parentheses ..

64x^2-((-9x^2-6x+6x+4))=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+4

Back to the equation:

-(-9x^2+4)


We get rid of parentheses

64x^2+9x^2-4=0

We add all the numbers together, and all the variables

73x^2-4=0

a = 73; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·73·(-4)
Δ = 1168
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:

$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{73}}{2*73}=\frac{0-4\sqrt{73}}{146}
=-\frac{4\sqrt{73}}{146}
=-\frac{2\sqrt{73}}{73}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{73}}{2*73}=\frac{0+4\sqrt{73}}{146}
=\frac{4\sqrt{73}}{146}
=\frac{2\sqrt{73}}{73}
$