Solve (x)=-3(x-1)^2+3

$$( x ) = - 3 ( x - 1 ) ^ { 2 } + 3$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x=-3\left(x^{2}-2x+1\right)+3$$

Use the distributive property to multiply -3 by x^{2}-2x+1.

$$x=-3x^{2}+6x-3+3$$

Add -3 and 3 to get 0.

$$x=-3x^{2}+6x$$

Add 3x^{2} to both sides.

$$x+3x^{2}=6x$$

Subtract 6x from both sides.

$$x+3x^{2}-6x=0$$

Combine x and -6x to get -5x.

$$-5x+3x^{2}=0$$

Factor out x.

$$x\left(-5+3x\right)=0$$

To find equation solutions, solve x=0 and -5+3x=0.

$$x=0 x=\frac{5}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x=-3\left(x^{2}-2x+1\right)+3$$

Use the distributive property to multiply -3 by x^{2}-2x+1.

$$x=-3x^{2}+6x-3+3$$

Add -3 and 3 to get 0.

$$x=-3x^{2}+6x$$

Add 3x^{2} to both sides.

$$x+3x^{2}=6x$$

Subtract 6x from both sides.

$$x+3x^{2}-6x=0$$

Combine x and -6x to get -5x.

$$-5x+3x^{2}=0$$

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.

$$3x^{2}-5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\times 3}$$

Take the square root of \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2\times 3}$$

The opposite of -5 is 5.

$$x=\frac{5±5}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{5±5}{6}$$

Now solve the equation x=\frac{5±5}{6} when ± is plus. Add 5 to 5.

$$x=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{5±5}{6} when ± is minus. Subtract 5 from 5.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=\frac{5}{3} x=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x=-3\left(x^{2}-2x+1\right)+3$$

Use the distributive property to multiply -3 by x^{2}-2x+1.

$$x=-3x^{2}+6x-3+3$$

Add -3 and 3 to get 0.

$$x=-3x^{2}+6x$$

Add 3x^{2} to both sides.

$$x+3x^{2}=6x$$

Subtract 6x from both sides.

$$x+3x^{2}-6x=0$$

Combine x and -6x to get -5x.

$$-5x+3x^{2}=0$$

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.

$$3x^{2}-5x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-5x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{5}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}-\frac{5}{3}x=0$$

Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=\left(-\frac{5}{6}\right)^{2}$$

Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{25}{36}$$

Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x-\frac{5}{6}=\frac{5}{6} x-\frac{5}{6}=-\frac{5}{6}$$

Add \frac{5}{6} to both sides of the equation.

$$x=\frac{5}{3} x=0$$