Solve y=(2x-3)^2/(9^2-1)

$$y = \frac { ( 2 x - 3 ) ^ { 2 } } { ( 9 ^ { 2 } - 1 ) }$$

Solve for y

$$y=\frac{\left(2x-3\right)^{2}}{80}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-3\right)^{2}.

$$y=\frac{4x^{2}-12x+9}{9^{2}-1}$$

Calculate 9 to the power of 2 and get 81.

$$y=\frac{4x^{2}-12x+9}{81-1}$$

Subtract 1 from 81 to get 80.

$$y=\frac{4x^{2}-12x+9}{80}$$

Divide each term of 4x^{2}-12x+9 by 80 to get \frac{1}{20}x^{2}-\frac{3}{20}x+\frac{9}{80}.

$$y=\frac{1}{20}x^{2}-\frac{3}{20}x+\frac{9}{80}$$

Solve for x (complex solution)

$$x=-2\sqrt{5y}+\frac{3}{2}$$

$$x=2\sqrt{5y}+\frac{3}{2}$$

Solve for x

$$x=-2\sqrt{5y}+\frac{3}{2}$$

$$x=2\sqrt{5y}+\frac{3}{2}\text{, }y\geq 0$$