Solve x:2(x-1)=4:3

$$x : 2 ( x - 1 ) = 4 : 3$$

Solve for x

$$x=\frac{\sqrt{105}}{6}+\frac{1}{2}\approx 2.207825128$$

$$x=-\frac{\sqrt{105}}{6}+\frac{1}{2}\approx -1.207825128$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Use the distributive property to multiply 3x by x-1.

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

Subtract 8 from both sides.

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

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

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-8\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-3\right)±\sqrt{9-12\left(-8\right)}}{2\times 3}$$

Multiply -12 times -8.

$$x=\frac{-\left(-3\right)±\sqrt{9+96}}{2\times 3}$$

Add 9 to 96.

$$x=\frac{-\left(-3\right)±\sqrt{105}}{2\times 3}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{105}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{3±\sqrt{105}}{6}$$

Now solve the equation x=\frac{3±\sqrt{105}}{6} when ± is plus. Add 3 to \sqrt{105}.

$$x=\frac{\sqrt{105}+3}{6}$$

Divide 3+\sqrt{105} by 6.

$$x=\frac{\sqrt{105}}{6}+\frac{1}{2}$$

Now solve the equation x=\frac{3±\sqrt{105}}{6} when ± is minus. Subtract \sqrt{105} from 3.

$$x=\frac{3-\sqrt{105}}{6}$$

Divide 3-\sqrt{105} by 6.

$$x=-\frac{\sqrt{105}}{6}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{105}}{6}+\frac{1}{2} x=-\frac{\sqrt{105}}{6}+\frac{1}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Use the distributive property to multiply 3x by x-1.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -3 by 3.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{8}{3}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{8}{3}+\frac{1}{4}$$

Add \frac{8}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{35}{12}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{35}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{35}{12}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{105}}{6} x-\frac{1}{2}=-\frac{\sqrt{105}}{6}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{105}}{6}+\frac{1}{2} x=-\frac{\sqrt{105}}{6}+\frac{1}{2}$$