Solve (3x+2):4x=5:6

$$( 3 x + 2 ) : 4 x = 5 : 6$$

Solve for x

$$x=\frac{\sqrt{11}-1}{3}\approx 0.772208263$$

$$x=\frac{-\sqrt{11}-1}{3}\approx -1.43887493$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 4,6.

$$3\left(3x+2\right)x=10$$

Use the distributive property to multiply 3 by 3x+2.

$$\left(9x+6\right)x=10$$

Use the distributive property to multiply 9x+6 by x.

$$9x^{2}+6x=10$$

Subtract 10 from both sides.

$$9x^{2}+6x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 6 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 9\left(-10\right)}}{2\times 9}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 9\left(-10\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-6±\sqrt{36-36\left(-10\right)}}{2\times 9}$$

Multiply -36 times -10.

$$x=\frac{-6±\sqrt{36+360}}{2\times 9}$$

Add 36 to 360.

$$x=\frac{-6±\sqrt{396}}{2\times 9}$$

Take the square root of 396.

$$x=\frac{-6±6\sqrt{11}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-6±6\sqrt{11}}{18}$$

Now solve the equation x=\frac{-6±6\sqrt{11}}{18} when ± is plus. Add -6 to 6\sqrt{11}.

$$x=\frac{6\sqrt{11}-6}{18}$$

Divide -6+6\sqrt{11} by 18.

$$x=\frac{\sqrt{11}-1}{3}$$

Now solve the equation x=\frac{-6±6\sqrt{11}}{18} when ± is minus. Subtract 6\sqrt{11} from -6.

$$x=\frac{-6\sqrt{11}-6}{18}$$

Divide -6-6\sqrt{11} by 18.

$$x=\frac{-\sqrt{11}-1}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{11}-1}{3} x=\frac{-\sqrt{11}-1}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 12, the least common multiple of 4,6.

$$3\left(3x+2\right)x=10$$

Use the distributive property to multiply 3 by 3x+2.

$$\left(9x+6\right)x=10$$

Use the distributive property to multiply 9x+6 by x.

$$9x^{2}+6x=10$$

Divide both sides by 9.

$$\frac{9x^{2}+6x}{9}=\frac{10}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{6}{9}x=\frac{10}{9}$$

Reduce the fraction \frac{6}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{2}{3}x=\frac{10}{9}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{10}{9}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{10+1}{9}$$

Add \frac{10}{9} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{11}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{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}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{11}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{11}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{\sqrt{11}}{3} x+\frac{1}{3}=-\frac{\sqrt{11}}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{\sqrt{11}-1}{3} x=\frac{-\sqrt{11}-1}{3}$$