Solve left(2x-5right)^2-left(x-6right)^2=0

$${ \left(2x-5 \right) }^{ 2 } - { \left(x-6 \right) }^{ 2 } =0$$

Solve for x

$$x=-1$$

$$x = \frac{11}{3} = 3\frac{2}{3} \approx 3.666666667$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$4x^{2}-20x+25-\left(x-6\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$4x^{2}-20x+25-\left(x^{2}-12x+36\right)=0$$

To find the opposite of x^{2}-12x+36, find the opposite of each term.

$$4x^{2}-20x+25-x^{2}+12x-36=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-20x+25+12x-36=0$$

Combine -20x and 12x to get -8x.

$$3x^{2}-8x+25-36=0$$

Subtract 36 from 25 to get -11.

$$3x^{2}-8x-11=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-11. To find a and b, set up a system to be solved.

$$a+b=-8 ab=3\left(-11\right)=-33$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -33.

$$1,-33 3,-11$$

Calculate the sum for each pair.

$$1-33=-32 3-11=-8$$

The solution is the pair that gives sum -8.

$$a=-11 b=3$$

Rewrite 3x^{2}-8x-11 as \left(3x^{2}-11x\right)+\left(3x-11\right).

$$\left(3x^{2}-11x\right)+\left(3x-11\right)$$

Factor out x in 3x^{2}-11x.

$$x\left(3x-11\right)+3x-11$$

Factor out common term 3x-11 by using distributive property.

$$\left(3x-11\right)\left(x+1\right)$$

To find equation solutions, solve 3x-11=0 and x+1=0.

$$x=\frac{11}{3} x=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$4x^{2}-20x+25-\left(x-6\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$4x^{2}-20x+25-\left(x^{2}-12x+36\right)=0$$

To find the opposite of x^{2}-12x+36, find the opposite of each term.

$$4x^{2}-20x+25-x^{2}+12x-36=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-20x+25+12x-36=0$$

Combine -20x and 12x to get -8x.

$$3x^{2}-8x+25-36=0$$

Subtract 36 from 25 to get -11.

$$3x^{2}-8x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -8 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\left(-11\right)}}{2\times 3}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3\left(-11\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12\left(-11\right)}}{2\times 3}$$

Multiply -12 times -11.

$$x=\frac{-\left(-8\right)±\sqrt{64+132}}{2\times 3}$$

Add 64 to 132.

$$x=\frac{-\left(-8\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$x=\frac{-\left(-8\right)±14}{2\times 3}$$

The opposite of -8 is 8.

$$x=\frac{8±14}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{8±14}{6}$$

Now solve the equation x=\frac{8±14}{6} when ± is plus. Add 8 to 14.

$$x=\frac{22}{6}$$

Reduce the fraction \frac{22}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{11}{3}$$

Now solve the equation x=\frac{8±14}{6} when ± is minus. Subtract 14 from 8.

$$x=-\frac{6}{6}$$

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

$$x=\frac{11}{3} x=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$4x^{2}-20x+25-\left(x-6\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$4x^{2}-20x+25-\left(x^{2}-12x+36\right)=0$$

To find the opposite of x^{2}-12x+36, find the opposite of each term.

$$4x^{2}-20x+25-x^{2}+12x-36=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-20x+25+12x-36=0$$

Combine -20x and 12x to get -8x.

$$3x^{2}-8x+25-36=0$$

Subtract 36 from 25 to get -11.

$$3x^{2}-8x-11=0$$

Add 11 to both sides. Anything plus zero gives itself.

$$3x^{2}-8x=11$$

Divide both sides by 3.

$$\frac{3x^{2}-8x}{3}=\frac{11}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{8}{3}x=\frac{11}{3}$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=\frac{11}{3}+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{11}{3}+\frac{16}{9}$$

Add \frac{11}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{49}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x-\frac{4}{3}=\frac{7}{3} x-\frac{4}{3}=-\frac{7}{3}$$

Add \frac{4}{3} to both sides of the equation.

$$x=\frac{11}{3} x=-1$$