Solve x=-3(x-2)^2+4.Ee

$$x = - 3 ( x - 2 ) ^ { 2 } + 4 . Ee$$

Solve for x

$$x=1$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

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}.

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

Use the distributive property to multiply -3 by x^{2}-4x+4.

$$x=-3x^{2}+12x-12+4$$

Add -12 and 4 to get -8.

$$x=-3x^{2}+12x-8$$

Add 3x^{2} to both sides.

$$x+3x^{2}=12x-8$$

Subtract 12x from both sides.

$$x+3x^{2}-12x=-8$$

Combine x and -12x to get -11x.

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

Add 8 to both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-11 ab=3\times 8=24$$

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 24.

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

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -11.

$$a=-8 b=-3$$

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

$$\left(3x^{2}-8x\right)+\left(-3x+8\right)$$

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

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

Factor out common term 3x-8 by using distributive property.

$$\left(3x-8\right)\left(x-1\right)$$

To find equation solutions, solve 3x-8=0 and x-1=0.

$$x=\frac{8}{3} x=1$$

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}.

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

Use the distributive property to multiply -3 by x^{2}-4x+4.

$$x=-3x^{2}+12x-12+4$$

Add -12 and 4 to get -8.

$$x=-3x^{2}+12x-8$$

Add 3x^{2} to both sides.

$$x+3x^{2}=12x-8$$

Subtract 12x from both sides.

$$x+3x^{2}-12x=-8$$

Combine x and -12x to get -11x.

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

Add 8 to both sides.

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

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.

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

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

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

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 3\times 8}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-11\right)±\sqrt{121-12\times 8}}{2\times 3}$$

Multiply -12 times 8.

$$x=\frac{-\left(-11\right)±\sqrt{121-96}}{2\times 3}$$

Add 121 to -96.

$$x=\frac{-\left(-11\right)±\sqrt{25}}{2\times 3}$$

Take the square root of 25.

$$x=\frac{-\left(-11\right)±5}{2\times 3}$$

The opposite of -11 is 11.

$$x=\frac{11±5}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{11±5}{6}$$

Now solve the equation x=\frac{11±5}{6} when ± is plus. Add 11 to 5.

$$x=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{11±5}{6} when ± is minus. Subtract 5 from 11.

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

Divide 6 by 6.

$$x=1$$

The equation is now solved.

$$x=\frac{8}{3} x=1$$

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}.

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

Use the distributive property to multiply -3 by x^{2}-4x+4.

$$x=-3x^{2}+12x-12+4$$

Add -12 and 4 to get -8.

$$x=-3x^{2}+12x-8$$

Add 3x^{2} to both sides.

$$x+3x^{2}=12x-8$$

Subtract 12x from both sides.

$$x+3x^{2}-12x=-8$$

Combine x and -12x to get -11x.

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

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.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-\frac{8}{3}+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-\frac{8}{3}+\frac{121}{36}$$

Add -\frac{8}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{25}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{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}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{5}{6} x-\frac{11}{6}=-\frac{5}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$x=\frac{8}{3} x=1$$