Solve 2(x-2)^2=7x-5

$$2 ( x - 2 ) ^ { 2 } = 7 x - 5$$

Solve for x

$$x=1$$

$$x = \frac{13}{2} = 6\frac{1}{2} = 6.5$$

Steps Using Factoring By Grouping

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

$$2\left(x^{2}-4x+4\right)=7x-5$$

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

$$2x^{2}-8x+8=7x-5$$

Subtract 7x from both sides.

$$2x^{2}-8x+8-7x=-5$$

Combine -8x and -7x to get -15x.

$$2x^{2}-15x+8=-5$$

Add 5 to both sides.

$$2x^{2}-15x+8+5=0$$

Add 8 and 5 to get 13.

$$2x^{2}-15x+13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+13. To find a and b, set up a system to be solved.

$$a+b=-15 ab=2\times 13=26$$

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 26.

$$-1,-26 -2,-13$$

Calculate the sum for each pair.

$$-1-26=-27 -2-13=-15$$

The solution is the pair that gives sum -15.

$$a=-13 b=-2$$

Rewrite 2x^{2}-15x+13 as \left(2x^{2}-13x\right)+\left(-2x+13\right).

$$\left(2x^{2}-13x\right)+\left(-2x+13\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(2x-13\right)-\left(2x-13\right)$$

Factor out common term 2x-13 by using distributive property.

$$\left(2x-13\right)\left(x-1\right)$$

To find equation solutions, solve 2x-13=0 and x-1=0.

$$x=\frac{13}{2} x=1$$

Steps Using the Quadratic Formula

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

$$2\left(x^{2}-4x+4\right)=7x-5$$

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

$$2x^{2}-8x+8=7x-5$$

Subtract 7x from both sides.

$$2x^{2}-8x+8-7x=-5$$

Combine -8x and -7x to get -15x.

$$2x^{2}-15x+8=-5$$

Add 5 to both sides.

$$2x^{2}-15x+8+5=0$$

Add 8 and 5 to get 13.

$$2x^{2}-15x+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -15 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 2\times 13}}{2\times 2}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 2\times 13}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-15\right)±\sqrt{225-8\times 13}}{2\times 2}$$

Multiply -8 times 13.

$$x=\frac{-\left(-15\right)±\sqrt{225-104}}{2\times 2}$$

Add 225 to -104.

$$x=\frac{-\left(-15\right)±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-\left(-15\right)±11}{2\times 2}$$

The opposite of -15 is 15.

$$x=\frac{15±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{15±11}{4}$$

Now solve the equation x=\frac{15±11}{4} when ± is plus. Add 15 to 11.

$$x=\frac{26}{4}$$

Reduce the fraction \frac{26}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{2}$$

Now solve the equation x=\frac{15±11}{4} when ± is minus. Subtract 11 from 15.

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

Divide 4 by 4.

$$x=1$$

The equation is now solved.

$$x=\frac{13}{2} x=1$$

Steps for Completing the Square

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

$$2\left(x^{2}-4x+4\right)=7x-5$$

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

$$2x^{2}-8x+8=7x-5$$

Subtract 7x from both sides.

$$2x^{2}-8x+8-7x=-5$$

Combine -8x and -7x to get -15x.

$$2x^{2}-15x+8=-5$$

Subtract 8 from both sides.

$$2x^{2}-15x=-5-8$$

Subtract 8 from -5 to get -13.

$$2x^{2}-15x=-13$$

Divide both sides by 2.

$$\frac{2x^{2}-15x}{2}=-\frac{13}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{15}{2}x=-\frac{13}{2}$$

Divide -\frac{15}{2}, the coefficient of the x term, by 2 to get -\frac{15}{4}. Then add the square of -\frac{15}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{15}{2}x+\left(-\frac{15}{4}\right)^{2}=-\frac{13}{2}+\left(-\frac{15}{4}\right)^{2}$$

Square -\frac{15}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=-\frac{13}{2}+\frac{225}{16}$$

Add -\frac{13}{2} to \frac{225}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{121}{16}$$

Factor x^{2}-\frac{15}{2}x+\frac{225}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{15}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x-\frac{15}{4}=\frac{11}{4} x-\frac{15}{4}=-\frac{11}{4}$$

Add \frac{15}{4} to both sides of the equation.

$$x=\frac{13}{2} x=1$$