Solve (x^2-5/9):4-sqrt{25/81}=3-7*1/3

$$( x ^ { 2 } - \frac { 5 } { 9 } ) : 4 - \sqrt { \frac { 25 } { 81 } } = 3 - 7 \cdot \frac { 1 } { 3 }$$

Solve for x

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

Steps Using Difference of Squares

Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3\left(x^{2}-\frac{5}{9}\right)-12\sqrt{\frac{25}{81}}=36-84\times \frac{1}{3}$$

Use the distributive property to multiply 3 by x^{2}-\frac{5}{9}.

$$3x^{2}-\frac{5}{3}-12\sqrt{\frac{25}{81}}=36-84\times \frac{1}{3}$$

Rewrite the square root of the division \frac{25}{81} as the division of square roots \frac{\sqrt{25}}{\sqrt{81}}. Take the square root of both numerator and denominator.

$$3x^{2}-\frac{5}{3}-12\times \frac{5}{9}=36-84\times \frac{1}{3}$$

Multiply -12 and \frac{5}{9} to get -\frac{20}{3}.

$$3x^{2}-\frac{5}{3}-\frac{20}{3}=36-84\times \frac{1}{3}$$

Subtract \frac{20}{3} from -\frac{5}{3} to get -\frac{25}{3}.

$$3x^{2}-\frac{25}{3}=36-84\times \frac{1}{3}$$

Multiply 84 and \frac{1}{3} to get 28.

$$3x^{2}-\frac{25}{3}=36-28$$

Subtract 28 from 36 to get 8.

$$3x^{2}-\frac{25}{3}=8$$

Subtract 8 from both sides.

$$3x^{2}-\frac{25}{3}-8=0$$

Subtract 8 from -\frac{25}{3} to get -\frac{49}{3}.

$$3x^{2}-\frac{49}{3}=0$$

Multiply both sides by 3.

$$9x^{2}-49=0$$

Consider 9x^{2}-49. Rewrite 9x^{2}-49 as \left(3x\right)^{2}-7^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(3x-7\right)\left(3x+7\right)=0$$

To find equation solutions, solve 3x-7=0 and 3x+7=0.

$$x=\frac{7}{3} x=-\frac{7}{3}$$

Steps by Finding Square Root

Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3\left(x^{2}-\frac{5}{9}\right)-12\sqrt{\frac{25}{81}}=36-84\times \frac{1}{3}$$

Use the distributive property to multiply 3 by x^{2}-\frac{5}{9}.

$$3x^{2}-\frac{5}{3}-12\sqrt{\frac{25}{81}}=36-84\times \frac{1}{3}$$

Rewrite the square root of the division \frac{25}{81} as the division of square roots \frac{\sqrt{25}}{\sqrt{81}}. Take the square root of both numerator and denominator.

$$3x^{2}-\frac{5}{3}-12\times \frac{5}{9}=36-84\times \frac{1}{3}$$

Multiply -12 and \frac{5}{9} to get -\frac{20}{3}.

$$3x^{2}-\frac{5}{3}-\frac{20}{3}=36-84\times \frac{1}{3}$$

Subtract \frac{20}{3} from -\frac{5}{3} to get -\frac{25}{3}.

$$3x^{2}-\frac{25}{3}=36-84\times \frac{1}{3}$$

Multiply 84 and \frac{1}{3} to get 28.

$$3x^{2}-\frac{25}{3}=36-28$$

Subtract 28 from 36 to get 8.

$$3x^{2}-\frac{25}{3}=8$$

Add \frac{25}{3} to both sides.

$$3x^{2}=8+\frac{25}{3}$$

Add 8 and \frac{25}{3} to get \frac{49}{3}.

$$3x^{2}=\frac{49}{3}$$

Divide both sides by 3.

$$x^{2}=\frac{\frac{49}{3}}{3}$$

Express \frac{\frac{49}{3}}{3} as a single fraction.

$$x^{2}=\frac{49}{3\times 3}$$

Multiply 3 and 3 to get 9.

$$x^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$x=\frac{7}{3} x=-\frac{7}{3}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3\left(x^{2}-\frac{5}{9}\right)-12\sqrt{\frac{25}{81}}=36-84\times \frac{1}{3}$$

Use the distributive property to multiply 3 by x^{2}-\frac{5}{9}.

$$3x^{2}-\frac{5}{3}-12\sqrt{\frac{25}{81}}=36-84\times \frac{1}{3}$$

Rewrite the square root of the division \frac{25}{81} as the division of square roots \frac{\sqrt{25}}{\sqrt{81}}. Take the square root of both numerator and denominator.

$$3x^{2}-\frac{5}{3}-12\times \frac{5}{9}=36-84\times \frac{1}{3}$$

Multiply -12 and \frac{5}{9} to get -\frac{20}{3}.

$$3x^{2}-\frac{5}{3}-\frac{20}{3}=36-84\times \frac{1}{3}$$

Subtract \frac{20}{3} from -\frac{5}{3} to get -\frac{25}{3}.

$$3x^{2}-\frac{25}{3}=36-84\times \frac{1}{3}$$

Multiply 84 and \frac{1}{3} to get 28.

$$3x^{2}-\frac{25}{3}=36-28$$

Subtract 28 from 36 to get 8.

$$3x^{2}-\frac{25}{3}=8$$

Subtract 8 from both sides.

$$3x^{2}-\frac{25}{3}-8=0$$

Subtract 8 from -\frac{25}{3} to get -\frac{49}{3}.

$$3x^{2}-\frac{49}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -\frac{49}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-\frac{49}{3}\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-\frac{49}{3}\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-\frac{49}{3}\right)}}{2\times 3}$$

Multiply -12 times -\frac{49}{3}.

$$x=\frac{0±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$x=\frac{0±14}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±14}{6}$$

Now solve the equation x=\frac{0±14}{6} when ± is plus. Reduce the fraction \frac{14}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{3}$$

Now solve the equation x=\frac{0±14}{6} when ± is minus. Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{3}$$

The equation is now solved.

$$x=\frac{7}{3} x=-\frac{7}{3}$$