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

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

Solve for x

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

$$x=2$$

Steps Using Factoring By Grouping

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

$$x=2\left(x^{2}-6x+9\right)$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$x=2x^{2}-12x+18$$

Subtract 2x^{2} from both sides.

$$x-2x^{2}=-12x+18$$

Add 12x to both sides.

$$x-2x^{2}+12x=18$$

Combine x and 12x to get 13x.

$$13x-2x^{2}=18$$

Subtract 18 from both sides.

$$13x-2x^{2}-18=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}+13x-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=13 ab=-2\left(-18\right)=36$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 13.

$$a=9 b=4$$

Rewrite -2x^{2}+13x-18 as \left(-2x^{2}+9x\right)+\left(4x-18\right).

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

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

$$-x\left(2x-9\right)+2\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(-x+2\right)$$

To find equation solutions, solve 2x-9=0 and -x+2=0.

$$x=\frac{9}{2} x=2$$

Steps Using the Quadratic Formula

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

$$x=2\left(x^{2}-6x+9\right)$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$x=2x^{2}-12x+18$$

Subtract 2x^{2} from both sides.

$$x-2x^{2}=-12x+18$$

Add 12x to both sides.

$$x-2x^{2}+12x=18$$

Combine x and 12x to get 13x.

$$13x-2x^{2}=18$$

Subtract 18 from both sides.

$$13x-2x^{2}-18=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}+13x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 13 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-13±\sqrt{169+8\left(-18\right)}}{2\left(-2\right)}$$

Multiply 8 times -18.

$$x=\frac{-13±\sqrt{169-144}}{2\left(-2\right)}$$

Add 169 to -144.

$$x=\frac{-13±\sqrt{25}}{2\left(-2\right)}$$

Take the square root of 25.

$$x=\frac{-13±5}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-13±5}{-4}$$

Now solve the equation x=\frac{-13±5}{-4} when ± is plus. Add -13 to 5.

$$x=-\frac{8}{-4}$$

Divide -8 by -4.

$$x=2$$

Now solve the equation x=\frac{-13±5}{-4} when ± is minus. Subtract 5 from -13.

$$x=-\frac{18}{-4}$$

Reduce the fraction \frac{-18}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

The equation is now solved.

$$x=2 x=\frac{9}{2}$$

Steps for Completing the Square

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

$$x=2\left(x^{2}-6x+9\right)$$

Use the distributive property to multiply 2 by x^{2}-6x+9.

$$x=2x^{2}-12x+18$$

Subtract 2x^{2} from both sides.

$$x-2x^{2}=-12x+18$$

Add 12x to both sides.

$$x-2x^{2}+12x=18$$

Combine x and 12x to get 13x.

$$13x-2x^{2}=18$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}+13x=18$$

Divide both sides by -2.

$$\frac{-2x^{2}+13x}{-2}=\frac{18}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{13}{-2}x=\frac{18}{-2}$$

Divide 13 by -2.

$$x^{2}-\frac{13}{2}x=\frac{18}{-2}$$

Divide 18 by -2.

$$x^{2}-\frac{13}{2}x=-9$$

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=-9+\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=-9+\frac{169}{16}$$

Add -9 to \frac{169}{16}.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{13}{2}x+\frac{169}{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}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{5}{4} x-\frac{13}{4}=-\frac{5}{4}$$

Add \frac{13}{4} to both sides of the equation.

$$x=\frac{9}{2} x=2$$