Solve 3(2/3x-2/3)^2=3x^2+4x-7

$$3 ( \frac { 2 } { 3 } x - \frac { 2 } { 3 } ) ^ { 2 } = 3 x ^ { 2 } + 4 x - 7$$

Solve for x

$$x=1$$

$$x=-5$$

Steps Using the Quadratic Formula

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

$$3\left(\frac{4}{9}x^{2}-\frac{8}{9}x+\frac{4}{9}\right)=3x^{2}+4x-7$$

Use the distributive property to multiply 3 by \frac{4}{9}x^{2}-\frac{8}{9}x+\frac{4}{9}.

$$\frac{4}{3}x^{2}-\frac{8}{3}x+\frac{4}{3}=3x^{2}+4x-7$$

Subtract 3x^{2} from both sides.

$$\frac{4}{3}x^{2}-\frac{8}{3}x+\frac{4}{3}-3x^{2}=4x-7$$

Combine \frac{4}{3}x^{2} and -3x^{2} to get -\frac{5}{3}x^{2}.

$$-\frac{5}{3}x^{2}-\frac{8}{3}x+\frac{4}{3}=4x-7$$

Subtract 4x from both sides.

$$-\frac{5}{3}x^{2}-\frac{8}{3}x+\frac{4}{3}-4x=-7$$

Combine -\frac{8}{3}x and -4x to get -\frac{20}{3}x.

$$-\frac{5}{3}x^{2}-\frac{20}{3}x+\frac{4}{3}=-7$$

Add 7 to both sides.

$$-\frac{5}{3}x^{2}-\frac{20}{3}x+\frac{4}{3}+7=0$$

Add \frac{4}{3} and 7 to get \frac{25}{3}.

$$-\frac{5}{3}x^{2}-\frac{20}{3}x+\frac{25}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{5}{3} for a, -\frac{20}{3} for b, and \frac{25}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{\left(-\frac{20}{3}\right)^{2}-4\left(-\frac{5}{3}\right)\times \frac{25}{3}}}{2\left(-\frac{5}{3}\right)}$$

Square -\frac{20}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{\frac{400}{9}-4\left(-\frac{5}{3}\right)\times \frac{25}{3}}}{2\left(-\frac{5}{3}\right)}$$

Multiply -4 times -\frac{5}{3}.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{\frac{400}{9}+\frac{20}{3}\times \frac{25}{3}}}{2\left(-\frac{5}{3}\right)}$$

Multiply \frac{20}{3} times \frac{25}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{\frac{400+500}{9}}}{2\left(-\frac{5}{3}\right)}$$

Add \frac{400}{9} to \frac{500}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{20}{3}\right)±\sqrt{100}}{2\left(-\frac{5}{3}\right)}$$

Take the square root of 100.

$$x=\frac{-\left(-\frac{20}{3}\right)±10}{2\left(-\frac{5}{3}\right)}$$

The opposite of -\frac{20}{3} is \frac{20}{3}.

$$x=\frac{\frac{20}{3}±10}{2\left(-\frac{5}{3}\right)}$$

Multiply 2 times -\frac{5}{3}.

$$x=\frac{\frac{20}{3}±10}{-\frac{10}{3}}$$

Now solve the equation x=\frac{\frac{20}{3}±10}{-\frac{10}{3}} when ± is plus. Add \frac{20}{3} to 10.

$$x=\frac{\frac{50}{3}}{-\frac{10}{3}}$$

Divide \frac{50}{3} by -\frac{10}{3} by multiplying \frac{50}{3} by the reciprocal of -\frac{10}{3}.

$$x=-5$$

Now solve the equation x=\frac{\frac{20}{3}±10}{-\frac{10}{3}} when ± is minus. Subtract 10 from \frac{20}{3}.

$$x=-\frac{\frac{10}{3}}{-\frac{10}{3}}$$

Divide -\frac{10}{3} by -\frac{10}{3} by multiplying -\frac{10}{3} by the reciprocal of -\frac{10}{3}.

$$x=1$$

The equation is now solved.

$$x=-5 x=1$$

Steps for Completing the Square

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

$$3\left(\frac{4}{9}x^{2}-\frac{8}{9}x+\frac{4}{9}\right)=3x^{2}+4x-7$$

Use the distributive property to multiply 3 by \frac{4}{9}x^{2}-\frac{8}{9}x+\frac{4}{9}.

$$\frac{4}{3}x^{2}-\frac{8}{3}x+\frac{4}{3}=3x^{2}+4x-7$$

Subtract 3x^{2} from both sides.

$$\frac{4}{3}x^{2}-\frac{8}{3}x+\frac{4}{3}-3x^{2}=4x-7$$

Combine \frac{4}{3}x^{2} and -3x^{2} to get -\frac{5}{3}x^{2}.

$$-\frac{5}{3}x^{2}-\frac{8}{3}x+\frac{4}{3}=4x-7$$

Subtract 4x from both sides.

$$-\frac{5}{3}x^{2}-\frac{8}{3}x+\frac{4}{3}-4x=-7$$

Combine -\frac{8}{3}x and -4x to get -\frac{20}{3}x.

$$-\frac{5}{3}x^{2}-\frac{20}{3}x+\frac{4}{3}=-7$$

Subtract \frac{4}{3} from both sides.

$$-\frac{5}{3}x^{2}-\frac{20}{3}x=-7-\frac{4}{3}$$

Subtract \frac{4}{3} from -7 to get -\frac{25}{3}.

$$-\frac{5}{3}x^{2}-\frac{20}{3}x=-\frac{25}{3}$$

Divide both sides of the equation by -\frac{5}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{5}{3}x^{2}-\frac{20}{3}x}{-\frac{5}{3}}=-\frac{\frac{25}{3}}{-\frac{5}{3}}$$

Dividing by -\frac{5}{3} undoes the multiplication by -\frac{5}{3}.

$$x^{2}+\left(-\frac{\frac{20}{3}}{-\frac{5}{3}}\right)x=-\frac{\frac{25}{3}}{-\frac{5}{3}}$$

Divide -\frac{20}{3} by -\frac{5}{3} by multiplying -\frac{20}{3} by the reciprocal of -\frac{5}{3}.

$$x^{2}+4x=-\frac{\frac{25}{3}}{-\frac{5}{3}}$$

Divide -\frac{25}{3} by -\frac{5}{3} by multiplying -\frac{25}{3} by the reciprocal of -\frac{5}{3}.

$$x^{2}+4x=5$$

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

$$x^{2}+4x+2^{2}=5+2^{2}$$

Square 2.

$$x^{2}+4x+4=5+4$$

Add 5 to 4.

$$x^{2}+4x+4=9$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=3 x+2=-3$$

Subtract 2 from both sides of the equation.

$$x=1 x=-5$$