Solve x(2x-3)^2-(48x^2-x/4)=4(x-2)^3+1

$$x ( 2 x - 3 ) ^ { 2 } - ( \frac { 48 x ^ { 2 } - x } { 4 } ) = 4 ( x - 2 ) ^ { 3 } + 1$$

Solve for x

$$x=\frac{4}{5}=0.8$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4.

$$4x\left(2x-3\right)^{2}-\left(48x^{2}-x\right)=16\left(x-2\right)^{3}+4$$

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

$$4x\left(4x^{2}-12x+9\right)-\left(48x^{2}-x\right)=16\left(x-2\right)^{3}+4$$

Use the distributive property to multiply 4x by 4x^{2}-12x+9.

$$16x^{3}-48x^{2}+36x-\left(48x^{2}-x\right)=16\left(x-2\right)^{3}+4$$

To find the opposite of 48x^{2}-x, find the opposite of each term.

$$16x^{3}-48x^{2}+36x-48x^{2}+x=16\left(x-2\right)^{3}+4$$

Combine -48x^{2} and -48x^{2} to get -96x^{2}.

$$16x^{3}-96x^{2}+36x+x=16\left(x-2\right)^{3}+4$$

Combine 36x and x to get 37x.

$$16x^{3}-96x^{2}+37x=16\left(x-2\right)^{3}+4$$

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

$$16x^{3}-96x^{2}+37x=16\left(x^{3}-6x^{2}+12x-8\right)+4$$

Use the distributive property to multiply 16 by x^{3}-6x^{2}+12x-8.

$$16x^{3}-96x^{2}+37x=16x^{3}-96x^{2}+192x-128+4$$

Add -128 and 4 to get -124.

$$16x^{3}-96x^{2}+37x=16x^{3}-96x^{2}+192x-124$$

Subtract 16x^{3} from both sides.

$$16x^{3}-96x^{2}+37x-16x^{3}=-96x^{2}+192x-124$$

Combine 16x^{3} and -16x^{3} to get 0.

$$-96x^{2}+37x=-96x^{2}+192x-124$$

Add 96x^{2} to both sides.

$$-96x^{2}+37x+96x^{2}=192x-124$$

Combine -96x^{2} and 96x^{2} to get 0.

$$37x=192x-124$$

Subtract 192x from both sides.

$$37x-192x=-124$$

Combine 37x and -192x to get -155x.

$$-155x=-124$$

Divide both sides by -155.

$$x=\frac{-124}{-155}$$

Reduce the fraction \frac{-124}{-155} to lowest terms by extracting and canceling out -31.

$$x=\frac{4}{5}$$