Solve 4(3x-4)^2=(4x-3)^2

$$4 ( 3 x - 4 ) ^ { 2 } = ( 4 x - 3 ) ^ { 2 }$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x = \frac{11}{10} = 1\frac{1}{10} = 1.1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$4\left(9x^{2}-24x+16\right)=\left(4x-3\right)^{2}$$

Use the distributive property to multiply 4 by 9x^{2}-24x+16.

$$36x^{2}-96x+64=\left(4x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$36x^{2}-96x+64=16x^{2}-24x+9$$

Subtract 16x^{2} from both sides.

$$36x^{2}-96x+64-16x^{2}=-24x+9$$

Combine 36x^{2} and -16x^{2} to get 20x^{2}.

$$20x^{2}-96x+64=-24x+9$$

Add 24x to both sides.

$$20x^{2}-96x+64+24x=9$$

Combine -96x and 24x to get -72x.

$$20x^{2}-72x+64=9$$

Subtract 9 from both sides.

$$20x^{2}-72x+64-9=0$$

Subtract 9 from 64 to get 55.

$$20x^{2}-72x+55=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -72 for b, and 55 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-72\right)±\sqrt{\left(-72\right)^{2}-4\times 20\times 55}}{2\times 20}$$

Square -72.

$$x=\frac{-\left(-72\right)±\sqrt{5184-4\times 20\times 55}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-72\right)±\sqrt{5184-80\times 55}}{2\times 20}$$

Multiply -80 times 55.

$$x=\frac{-\left(-72\right)±\sqrt{5184-4400}}{2\times 20}$$

Add 5184 to -4400.

$$x=\frac{-\left(-72\right)±\sqrt{784}}{2\times 20}$$

Take the square root of 784.

$$x=\frac{-\left(-72\right)±28}{2\times 20}$$

The opposite of -72 is 72.

$$x=\frac{72±28}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{72±28}{40}$$

Now solve the equation x=\frac{72±28}{40} when ± is plus. Add 72 to 28.

$$x=\frac{100}{40}$$

Reduce the fraction \frac{100}{40} to lowest terms by extracting and canceling out 20.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{72±28}{40} when ± is minus. Subtract 28 from 72.

$$x=\frac{44}{40}$$

Reduce the fraction \frac{44}{40} to lowest terms by extracting and canceling out 4.

$$x=\frac{11}{10}$$

The equation is now solved.

$$x=\frac{5}{2} x=\frac{11}{10}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-4\right)^{2}.

$$4\left(9x^{2}-24x+16\right)=\left(4x-3\right)^{2}$$

Use the distributive property to multiply 4 by 9x^{2}-24x+16.

$$36x^{2}-96x+64=\left(4x-3\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-3\right)^{2}.

$$36x^{2}-96x+64=16x^{2}-24x+9$$

Subtract 16x^{2} from both sides.

$$36x^{2}-96x+64-16x^{2}=-24x+9$$

Combine 36x^{2} and -16x^{2} to get 20x^{2}.

$$20x^{2}-96x+64=-24x+9$$

Add 24x to both sides.

$$20x^{2}-96x+64+24x=9$$

Combine -96x and 24x to get -72x.

$$20x^{2}-72x+64=9$$

Subtract 64 from both sides.

$$20x^{2}-72x=9-64$$

Subtract 64 from 9 to get -55.

$$20x^{2}-72x=-55$$

Divide both sides by 20.

$$\frac{20x^{2}-72x}{20}=-\frac{55}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\left(-\frac{72}{20}\right)x=-\frac{55}{20}$$

Reduce the fraction \frac{-72}{20} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{18}{5}x=-\frac{55}{20}$$

Reduce the fraction \frac{-55}{20} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{18}{5}x=-\frac{11}{4}$$

Divide -\frac{18}{5}, the coefficient of the x term, by 2 to get -\frac{9}{5}. Then add the square of -\frac{9}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{18}{5}x+\left(-\frac{9}{5}\right)^{2}=-\frac{11}{4}+\left(-\frac{9}{5}\right)^{2}$$

Square -\frac{9}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=-\frac{11}{4}+\frac{81}{25}$$

Add -\frac{11}{4} to \frac{81}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{18}{5}x+\frac{81}{25}=\frac{49}{100}$$

Factor x^{2}-\frac{18}{5}x+\frac{81}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{5}\right)^{2}=\frac{49}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{5}\right)^{2}}=\sqrt{\frac{49}{100}}$$

Simplify.

$$x-\frac{9}{5}=\frac{7}{10} x-\frac{9}{5}=-\frac{7}{10}$$

Add \frac{9}{5} to both sides of the equation.

$$x=\frac{5}{2} x=\frac{11}{10}$$