Solve left(2x-5right)^2=left(x+3right)^2

$${ \left(2x-5 \right) }^{ 2 } = { \left(x+3 \right) }^{ 2 }$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=8$$

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}.

$$4x^{2}-20x+25=\left(x+3\right)^{2}$$

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

$$4x^{2}-20x+25=x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$4x^{2}-20x+25-x^{2}=6x+9$$

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

$$3x^{2}-20x+25=6x+9$$

Subtract 6x from both sides.

$$3x^{2}-20x+25-6x=9$$

Combine -20x and -6x to get -26x.

$$3x^{2}-26x+25=9$$

Subtract 9 from both sides.

$$3x^{2}-26x+25-9=0$$

Subtract 9 from 25 to get 16.

$$3x^{2}-26x+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=-26 ab=3\times 16=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -26.

$$a=-24 b=-2$$

Rewrite 3x^{2}-26x+16 as \left(3x^{2}-24x\right)+\left(-2x+16\right).

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

Factor out 3x in the first and -2 in the second group.

$$3x\left(x-8\right)-2\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(3x-2\right)$$

To find equation solutions, solve x-8=0 and 3x-2=0.

$$x=8 x=\frac{2}{3}$$

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}.

$$4x^{2}-20x+25=\left(x+3\right)^{2}$$

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

$$4x^{2}-20x+25=x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$4x^{2}-20x+25-x^{2}=6x+9$$

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

$$3x^{2}-20x+25=6x+9$$

Subtract 6x from both sides.

$$3x^{2}-20x+25-6x=9$$

Combine -20x and -6x to get -26x.

$$3x^{2}-26x+25=9$$

Subtract 9 from both sides.

$$3x^{2}-26x+25-9=0$$

Subtract 9 from 25 to get 16.

$$3x^{2}-26x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -26 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 3\times 16}}{2\times 3}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\times 3\times 16}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-26\right)±\sqrt{676-12\times 16}}{2\times 3}$$

Multiply -12 times 16.

$$x=\frac{-\left(-26\right)±\sqrt{676-192}}{2\times 3}$$

Add 676 to -192.

$$x=\frac{-\left(-26\right)±\sqrt{484}}{2\times 3}$$

Take the square root of 484.

$$x=\frac{-\left(-26\right)±22}{2\times 3}$$

The opposite of -26 is 26.

$$x=\frac{26±22}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{26±22}{6}$$

Now solve the equation x=\frac{26±22}{6} when ± is plus. Add 26 to 22.

$$x=\frac{48}{6}$$

Divide 48 by 6.

$$x=8$$

Now solve the equation x=\frac{26±22}{6} when ± is minus. Subtract 22 from 26.

$$x=\frac{4}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{3}$$

The equation is now solved.

$$x=8 x=\frac{2}{3}$$

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}.

$$4x^{2}-20x+25=\left(x+3\right)^{2}$$

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

$$4x^{2}-20x+25=x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$4x^{2}-20x+25-x^{2}=6x+9$$

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

$$3x^{2}-20x+25=6x+9$$

Subtract 6x from both sides.

$$3x^{2}-20x+25-6x=9$$

Combine -20x and -6x to get -26x.

$$3x^{2}-26x+25=9$$

Subtract 25 from both sides.

$$3x^{2}-26x=9-25$$

Subtract 25 from 9 to get -16.

$$3x^{2}-26x=-16$$

Divide both sides by 3.

$$\frac{3x^{2}-26x}{3}=-\frac{16}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{26}{3}x=-\frac{16}{3}$$

Divide -\frac{26}{3}, the coefficient of the x term, by 2 to get -\frac{13}{3}. Then add the square of -\frac{13}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{26}{3}x+\left(-\frac{13}{3}\right)^{2}=-\frac{16}{3}+\left(-\frac{13}{3}\right)^{2}$$

Square -\frac{13}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{26}{3}x+\frac{169}{9}=-\frac{16}{3}+\frac{169}{9}$$

Add -\frac{16}{3} to \frac{169}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{26}{3}x+\frac{169}{9}=\frac{121}{9}$$

Factor x^{2}-\frac{26}{3}x+\frac{169}{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}.

$$\left(x-\frac{13}{3}\right)^{2}=\frac{121}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{3}\right)^{2}}=\sqrt{\frac{121}{9}}$$

Simplify.

$$x-\frac{13}{3}=\frac{11}{3} x-\frac{13}{3}=-\frac{11}{3}$$

Add \frac{13}{3} to both sides of the equation.

$$x=8 x=\frac{2}{3}$$