Solve 4/3xleft(4+xright)=6/5left(x+4right)left(6/5left(4/3-3right)+10right)frac{1}{2}

$$\frac{ 4 }{ 3 } x \left( 4+x \right) = \frac{ 6 }{ 5 } \left( x+4 \right) \left( \frac{ 6 }{ 5 } \left( \frac{ 4 }{ 3 } -3 \right) +10 \right) \frac{ 1 }{ 2 }$$

Solve for x

$$x=-4$$

$$x = \frac{18}{5} = 3\frac{3}{5} = 3.6$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{4}{3}x by 4+x.

$$\frac{4}{3}x\times 4+\frac{4}{3}xx=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-3\right)+10\right)\times \frac{1}{2}$$

Multiply x and x to get x^{2}.

$$\frac{4}{3}x\times 4+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-3\right)+10\right)\times \frac{1}{2}$$

Express \frac{4}{3}\times 4 as a single fraction.

$$\frac{4\times 4}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-3\right)+10\right)\times \frac{1}{2}$$

Multiply 4 and 4 to get 16.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-3\right)+10\right)\times \frac{1}{2}$$

Convert 3 to fraction \frac{9}{3}.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-\frac{9}{3}\right)+10\right)\times \frac{1}{2}$$

Since \frac{4}{3} and \frac{9}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\times \frac{4-9}{3}+10\right)\times \frac{1}{2}$$

Subtract 9 from 4 to get -5.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(-\frac{5}{3}\right)+10\right)\times \frac{1}{2}$$

Multiply \frac{6}{5} times -\frac{5}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6\left(-5\right)}{5\times 3}+10\right)\times \frac{1}{2}$$

Do the multiplications in the fraction \frac{6\left(-5\right)}{5\times 3}.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{-30}{15}+10\right)\times \frac{1}{2}$$

Divide -30 by 15 to get -2.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(-2+10\right)\times \frac{1}{2}$$

Add -2 and 10 to get 8.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\times 8\times \frac{1}{2}$$

Express \frac{6}{5}\times 8 as a single fraction.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6\times 8}{5}\left(x+4\right)\times \frac{1}{2}$$

Multiply 6 and 8 to get 48.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{48}{5}\left(x+4\right)\times \frac{1}{2}$$

Multiply \frac{48}{5} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{48\times 1}{5\times 2}\left(x+4\right)$$

Do the multiplications in the fraction \frac{48\times 1}{5\times 2}.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{48}{10}\left(x+4\right)$$

Reduce the fraction \frac{48}{10} to lowest terms by extracting and canceling out 2.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{24}{5}\left(x+4\right)$$

Use the distributive property to multiply \frac{24}{5} by x+4.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{24}{5}x+\frac{24}{5}\times 4$$

Express \frac{24}{5}\times 4 as a single fraction.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{24}{5}x+\frac{24\times 4}{5}$$

Multiply 24 and 4 to get 96.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{24}{5}x+\frac{96}{5}$$

Subtract \frac{24}{5}x from both sides.

$$\frac{16}{3}x+\frac{4}{3}x^{2}-\frac{24}{5}x=\frac{96}{5}$$

Combine \frac{16}{3}x and -\frac{24}{5}x to get \frac{8}{15}x.

$$\frac{8}{15}x+\frac{4}{3}x^{2}=\frac{96}{5}$$

Subtract \frac{96}{5} from both sides.

$$\frac{8}{15}x+\frac{4}{3}x^{2}-\frac{96}{5}=0$$

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.

$$\frac{4}{3}x^{2}+\frac{8}{15}x-\frac{96}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{4}{3} for a, \frac{8}{15} for b, and -\frac{96}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{8}{15}±\sqrt{\left(\frac{8}{15}\right)^{2}-4\times \frac{4}{3}\left(-\frac{96}{5}\right)}}{2\times \frac{4}{3}}$$

Square \frac{8}{15} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{8}{15}±\sqrt{\frac{64}{225}-4\times \frac{4}{3}\left(-\frac{96}{5}\right)}}{2\times \frac{4}{3}}$$

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

$$x=\frac{-\frac{8}{15}±\sqrt{\frac{64}{225}-\frac{16}{3}\left(-\frac{96}{5}\right)}}{2\times \frac{4}{3}}$$

Multiply -\frac{16}{3} times -\frac{96}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{8}{15}±\sqrt{\frac{64}{225}+\frac{512}{5}}}{2\times \frac{4}{3}}$$

Add \frac{64}{225} to \frac{512}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{8}{15}±\sqrt{\frac{23104}{225}}}{2\times \frac{4}{3}}$$

Take the square root of \frac{23104}{225}.

$$x=\frac{-\frac{8}{15}±\frac{152}{15}}{2\times \frac{4}{3}}$$

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

$$x=\frac{-\frac{8}{15}±\frac{152}{15}}{\frac{8}{3}}$$

Now solve the equation x=\frac{-\frac{8}{15}±\frac{152}{15}}{\frac{8}{3}} when ± is plus. Add -\frac{8}{15} to \frac{152}{15} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{48}{5}}{\frac{8}{3}}$$

Divide \frac{48}{5} by \frac{8}{3} by multiplying \frac{48}{5} by the reciprocal of \frac{8}{3}.

$$x=\frac{18}{5}$$

Now solve the equation x=\frac{-\frac{8}{15}±\frac{152}{15}}{\frac{8}{3}} when ± is minus. Subtract \frac{152}{15} from -\frac{8}{15} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{32}{3}}{\frac{8}{3}}$$

Divide -\frac{32}{3} by \frac{8}{3} by multiplying -\frac{32}{3} by the reciprocal of \frac{8}{3}.

$$x=-4$$

The equation is now solved.

$$x=\frac{18}{5} x=-4$$

Steps for Completing the Square

Use the distributive property to multiply \frac{4}{3}x by 4+x.

$$\frac{4}{3}x\times 4+\frac{4}{3}xx=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-3\right)+10\right)\times \frac{1}{2}$$

Multiply x and x to get x^{2}.

$$\frac{4}{3}x\times 4+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-3\right)+10\right)\times \frac{1}{2}$$

Express \frac{4}{3}\times 4 as a single fraction.

$$\frac{4\times 4}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-3\right)+10\right)\times \frac{1}{2}$$

Multiply 4 and 4 to get 16.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-3\right)+10\right)\times \frac{1}{2}$$

Convert 3 to fraction \frac{9}{3}.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(\frac{4}{3}-\frac{9}{3}\right)+10\right)\times \frac{1}{2}$$

Since \frac{4}{3} and \frac{9}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\times \frac{4-9}{3}+10\right)\times \frac{1}{2}$$

Subtract 9 from 4 to get -5.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6}{5}\left(-\frac{5}{3}\right)+10\right)\times \frac{1}{2}$$

Multiply \frac{6}{5} times -\frac{5}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{6\left(-5\right)}{5\times 3}+10\right)\times \frac{1}{2}$$

Do the multiplications in the fraction \frac{6\left(-5\right)}{5\times 3}.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(\frac{-30}{15}+10\right)\times \frac{1}{2}$$

Divide -30 by 15 to get -2.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\left(-2+10\right)\times \frac{1}{2}$$

Add -2 and 10 to get 8.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6}{5}\left(x+4\right)\times 8\times \frac{1}{2}$$

Express \frac{6}{5}\times 8 as a single fraction.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{6\times 8}{5}\left(x+4\right)\times \frac{1}{2}$$

Multiply 6 and 8 to get 48.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{48}{5}\left(x+4\right)\times \frac{1}{2}$$

Multiply \frac{48}{5} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{48\times 1}{5\times 2}\left(x+4\right)$$

Do the multiplications in the fraction \frac{48\times 1}{5\times 2}.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{48}{10}\left(x+4\right)$$

Reduce the fraction \frac{48}{10} to lowest terms by extracting and canceling out 2.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{24}{5}\left(x+4\right)$$

Use the distributive property to multiply \frac{24}{5} by x+4.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{24}{5}x+\frac{24}{5}\times 4$$

Express \frac{24}{5}\times 4 as a single fraction.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{24}{5}x+\frac{24\times 4}{5}$$

Multiply 24 and 4 to get 96.

$$\frac{16}{3}x+\frac{4}{3}x^{2}=\frac{24}{5}x+\frac{96}{5}$$

Subtract \frac{24}{5}x from both sides.

$$\frac{16}{3}x+\frac{4}{3}x^{2}-\frac{24}{5}x=\frac{96}{5}$$

Combine \frac{16}{3}x and -\frac{24}{5}x to get \frac{8}{15}x.

$$\frac{8}{15}x+\frac{4}{3}x^{2}=\frac{96}{5}$$

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.

$$\frac{4}{3}x^{2}+\frac{8}{15}x=\frac{96}{5}$$

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

$$\frac{\frac{4}{3}x^{2}+\frac{8}{15}x}{\frac{4}{3}}=\frac{\frac{96}{5}}{\frac{4}{3}}$$

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

$$x^{2}+\frac{\frac{8}{15}}{\frac{4}{3}}x=\frac{\frac{96}{5}}{\frac{4}{3}}$$

Divide \frac{8}{15} by \frac{4}{3} by multiplying \frac{8}{15} by the reciprocal of \frac{4}{3}.

$$x^{2}+\frac{2}{5}x=\frac{\frac{96}{5}}{\frac{4}{3}}$$

Divide \frac{96}{5} by \frac{4}{3} by multiplying \frac{96}{5} by the reciprocal of \frac{4}{3}.

$$x^{2}+\frac{2}{5}x=\frac{72}{5}$$

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{5}x+\left(\frac{1}{5}\right)^{2}=\frac{72}{5}+\left(\frac{1}{5}\right)^{2}$$

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{72}{5}+\frac{1}{25}$$

Add \frac{72}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{361}{25}$$

Factor x^{2}+\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{5}\right)^{2}=\frac{361}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{\frac{361}{25}}$$

Simplify.

$$x+\frac{1}{5}=\frac{19}{5} x+\frac{1}{5}=-\frac{19}{5}$$

Subtract \frac{1}{5} from both sides of the equation.

$$x=\frac{18}{5} x=-4$$