Solve 4/3x^2-8x+12=0

$$\frac{ 4 }{ 3 } { x }^{ 2 } -8x+12=0$$

Solve for x

$$x=3$$

Steps Using the Quadratic Formula

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.

$$\frac{4}{3}x^{2}-8x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{4}{3} for a, -8 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times \frac{4}{3}\times 12}}{2\times \frac{4}{3}}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times \frac{4}{3}\times 12}}{2\times \frac{4}{3}}$$

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

$$x=\frac{-\left(-8\right)±\sqrt{64-\frac{16}{3}\times 12}}{2\times \frac{4}{3}}$$

Multiply -\frac{16}{3} times 12.

$$x=\frac{-\left(-8\right)±\sqrt{64-64}}{2\times \frac{4}{3}}$$

Add 64 to -64.

$$x=\frac{-\left(-8\right)±\sqrt{0}}{2\times \frac{4}{3}}$$

Take the square root of 0.

$$x=-\frac{-8}{2\times \frac{4}{3}}$$

The opposite of -8 is 8.

$$x=\frac{8}{2\times \frac{4}{3}}$$

Multiply 2 times \frac{4}{3}.

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

Divide 8 by \frac{8}{3} by multiplying 8 by the reciprocal of \frac{8}{3}.

$$x=3$$

Steps for Completing the Square

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.

$$\frac{4}{3}x^{2}-8x+12=0$$

Subtract 12 from both sides of the equation.

$$\frac{4}{3}x^{2}-8x+12-12=-12$$

Subtracting 12 from itself leaves 0.

$$\frac{4}{3}x^{2}-8x=-12$$

Divide both sides of the equation by \frac{4}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{4}{3}x^{2}-8x}{\frac{4}{3}}=-\frac{12}{\frac{4}{3}}$$

Dividing by \frac{4}{3} undoes the multiplication by \frac{4}{3}.

$$x^{2}+\left(-\frac{8}{\frac{4}{3}}\right)x=-\frac{12}{\frac{4}{3}}$$

Divide -8 by \frac{4}{3} by multiplying -8 by the reciprocal of \frac{4}{3}.

$$x^{2}-6x=-\frac{12}{\frac{4}{3}}$$

Divide -12 by \frac{4}{3} by multiplying -12 by the reciprocal of \frac{4}{3}.

$$x^{2}-6x=-9$$

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

$$x^{2}-6x+\left(-3\right)^{2}=-9+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-9+9$$

Add -9 to 9.

$$x^{2}-6x+9=0$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-3=0 x-3=0$$

Add 3 to both sides of the equation.

$$x=3 x=3$$

The equation is now solved. Solutions are the same.

$$x=3$$