Solve 0.5*(2-x)(2-2/3x)=5

$$0.5 \times (2-x)(2- \frac{ 2 }{ 3 } x)=5$$

Solve for x

$$x = \frac{\sqrt{61} + 5}{2} \approx 6.405124838$$

$$x=\frac{5-\sqrt{61}}{2}\approx -1.405124838$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 0.5 by 2-x.

$$\left(1-0.5x\right)\left(2-\frac{2}{3}x\right)=5$$

Use the distributive property to multiply 1-0.5x by 2-\frac{2}{3}x and combine like terms.

$$2-\frac{5}{3}x+\frac{1}{3}x^{2}=5$$

Subtract 5 from both sides.

$$2-\frac{5}{3}x+\frac{1}{3}x^{2}-5=0$$

Subtract 5 from 2 to get -3.

$$-3-\frac{5}{3}x+\frac{1}{3}x^{2}=0$$

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.

$$\frac{1}{3}x^{2}-\frac{5}{3}x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{3} for a, -\frac{5}{3} for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\left(-\frac{5}{3}\right)^{2}-4\times \frac{1}{3}\left(-3\right)}}{2\times \frac{1}{3}}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{25}{9}-4\times \frac{1}{3}\left(-3\right)}}{2\times \frac{1}{3}}$$

Multiply -4 times \frac{1}{3}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{25}{9}-\frac{4}{3}\left(-3\right)}}{2\times \frac{1}{3}}$$

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

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{25}{9}+4}}{2\times \frac{1}{3}}$$

Add \frac{25}{9} to 4.

$$x=\frac{-\left(-\frac{5}{3}\right)±\sqrt{\frac{61}{9}}}{2\times \frac{1}{3}}$$

Take the square root of \frac{61}{9}.

$$x=\frac{-\left(-\frac{5}{3}\right)±\frac{\sqrt{61}}{3}}{2\times \frac{1}{3}}$$

The opposite of -\frac{5}{3} is \frac{5}{3}.

$$x=\frac{\frac{5}{3}±\frac{\sqrt{61}}{3}}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$x=\frac{\frac{5}{3}±\frac{\sqrt{61}}{3}}{\frac{2}{3}}$$

Now solve the equation x=\frac{\frac{5}{3}±\frac{\sqrt{61}}{3}}{\frac{2}{3}} when ± is plus. Add \frac{5}{3} to \frac{\sqrt{61}}{3}.

$$x=\frac{\sqrt{61}+5}{\frac{2}{3}\times 3}$$

Divide \frac{5+\sqrt{61}}{3} by \frac{2}{3} by multiplying \frac{5+\sqrt{61}}{3} by the reciprocal of \frac{2}{3}.

$$x=\frac{\sqrt{61}+5}{2}$$

Now solve the equation x=\frac{\frac{5}{3}±\frac{\sqrt{61}}{3}}{\frac{2}{3}} when ± is minus. Subtract \frac{\sqrt{61}}{3} from \frac{5}{3}.

$$x=\frac{5-\sqrt{61}}{\frac{2}{3}\times 3}$$

Divide \frac{5-\sqrt{61}}{3} by \frac{2}{3} by multiplying \frac{5-\sqrt{61}}{3} by the reciprocal of \frac{2}{3}.

$$x=\frac{5-\sqrt{61}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{61}+5}{2} x=\frac{5-\sqrt{61}}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 0.5 by 2-x.

$$\left(1-0.5x\right)\left(2-\frac{2}{3}x\right)=5$$

Use the distributive property to multiply 1-0.5x by 2-\frac{2}{3}x and combine like terms.

$$2-\frac{5}{3}x+\frac{1}{3}x^{2}=5$$

Subtract 2 from both sides.

$$-\frac{5}{3}x+\frac{1}{3}x^{2}=5-2$$

Subtract 2 from 5 to get 3.

$$-\frac{5}{3}x+\frac{1}{3}x^{2}=3$$

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.

$$\frac{1}{3}x^{2}-\frac{5}{3}x=3$$

Multiply both sides by 3.

$$\frac{\frac{1}{3}x^{2}-\frac{5}{3}x}{\frac{1}{3}}=\frac{3}{\frac{1}{3}}$$

Dividing by \frac{1}{3} undoes the multiplication by \frac{1}{3}.

$$x^{2}+\left(-\frac{\frac{5}{3}}{\frac{1}{3}}\right)x=\frac{3}{\frac{1}{3}}$$

Divide -\frac{5}{3} by \frac{1}{3} by multiplying -\frac{5}{3} by the reciprocal of \frac{1}{3}.

$$x^{2}-5x=\frac{3}{\frac{1}{3}}$$

Divide 3 by \frac{1}{3} by multiplying 3 by the reciprocal of \frac{1}{3}.

$$x^{2}-5x=9$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=9+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=9+\frac{25}{4}$$

Add 9 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{61}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{61}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{61}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{61}}{2} x-\frac{5}{2}=-\frac{\sqrt{61}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{61}+5}{2} x=\frac{5-\sqrt{61}}{2}$$