Solve 16/5x^2=(36/5-16/25x)(36-16/5x)

$$\frac { 16 } { 5 } x ^ { 2 } = ( \frac { 36 } { 5 } - \frac { 16 } { 25 } x ) ( 36 - \frac { 16 } { 5 } x )$$

Solve for x

$$x=-45$$

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{36}{5}-\frac{16}{25}x by 36-\frac{16}{5}x and combine like terms.

$$\frac{16}{5}x^{2}=\frac{1296}{5}-\frac{1152}{25}x+\frac{256}{125}x^{2}$$

Subtract \frac{1296}{5} from both sides.

$$\frac{16}{5}x^{2}-\frac{1296}{5}=-\frac{1152}{25}x+\frac{256}{125}x^{2}$$

Add \frac{1152}{25}x to both sides.

$$\frac{16}{5}x^{2}-\frac{1296}{5}+\frac{1152}{25}x=\frac{256}{125}x^{2}$$

Subtract \frac{256}{125}x^{2} from both sides.

$$\frac{16}{5}x^{2}-\frac{1296}{5}+\frac{1152}{25}x-\frac{256}{125}x^{2}=0$$

Combine \frac{16}{5}x^{2} and -\frac{256}{125}x^{2} to get \frac{144}{125}x^{2}.

$$\frac{144}{125}x^{2}-\frac{1296}{5}+\frac{1152}{25}x=0$$

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.

$$\frac{144}{125}x^{2}+\frac{1152}{25}x-\frac{1296}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{144}{125} for a, \frac{1152}{25} for b, and -\frac{1296}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1152}{25}±\sqrt{\left(\frac{1152}{25}\right)^{2}-4\times \frac{144}{125}\left(-\frac{1296}{5}\right)}}{2\times \frac{144}{125}}$$

Square \frac{1152}{25} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{1152}{25}±\sqrt{\frac{1327104}{625}-4\times \frac{144}{125}\left(-\frac{1296}{5}\right)}}{2\times \frac{144}{125}}$$

Multiply -4 times \frac{144}{125}.

$$x=\frac{-\frac{1152}{25}±\sqrt{\frac{1327104}{625}-\frac{576}{125}\left(-\frac{1296}{5}\right)}}{2\times \frac{144}{125}}$$

Multiply -\frac{576}{125} times -\frac{1296}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{1152}{25}±\sqrt{\frac{1327104+746496}{625}}}{2\times \frac{144}{125}}$$

Add \frac{1327104}{625} to \frac{746496}{625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{1152}{25}±\sqrt{\frac{82944}{25}}}{2\times \frac{144}{125}}$$

Take the square root of \frac{82944}{25}.

$$x=\frac{-\frac{1152}{25}±\frac{288}{5}}{2\times \frac{144}{125}}$$

Multiply 2 times \frac{144}{125}.

$$x=\frac{-\frac{1152}{25}±\frac{288}{5}}{\frac{288}{125}}$$

Now solve the equation x=\frac{-\frac{1152}{25}±\frac{288}{5}}{\frac{288}{125}} when ± is plus. Add -\frac{1152}{25} to \frac{288}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{288}{25}}{\frac{288}{125}}$$

Divide \frac{288}{25} by \frac{288}{125} by multiplying \frac{288}{25} by the reciprocal of \frac{288}{125}.

$$x=5$$

Now solve the equation x=\frac{-\frac{1152}{25}±\frac{288}{5}}{\frac{288}{125}} when ± is minus. Subtract \frac{288}{5} from -\frac{1152}{25} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{2592}{25}}{\frac{288}{125}}$$

Divide -\frac{2592}{25} by \frac{288}{125} by multiplying -\frac{2592}{25} by the reciprocal of \frac{288}{125}.

$$x=-45$$

The equation is now solved.

$$x=5 x=-45$$

Steps for Completing the Square

Use the distributive property to multiply \frac{36}{5}-\frac{16}{25}x by 36-\frac{16}{5}x and combine like terms.

$$\frac{16}{5}x^{2}=\frac{1296}{5}-\frac{1152}{25}x+\frac{256}{125}x^{2}$$

Add \frac{1152}{25}x to both sides.

$$\frac{16}{5}x^{2}+\frac{1152}{25}x=\frac{1296}{5}+\frac{256}{125}x^{2}$$

Subtract \frac{256}{125}x^{2} from both sides.

$$\frac{16}{5}x^{2}+\frac{1152}{25}x-\frac{256}{125}x^{2}=\frac{1296}{5}$$

Combine \frac{16}{5}x^{2} and -\frac{256}{125}x^{2} to get \frac{144}{125}x^{2}.

$$\frac{144}{125}x^{2}+\frac{1152}{25}x=\frac{1296}{5}$$

Divide both sides of the equation by \frac{144}{125}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{144}{125}x^{2}+\frac{1152}{25}x}{\frac{144}{125}}=\frac{\frac{1296}{5}}{\frac{144}{125}}$$

Dividing by \frac{144}{125} undoes the multiplication by \frac{144}{125}.

$$x^{2}+\frac{\frac{1152}{25}}{\frac{144}{125}}x=\frac{\frac{1296}{5}}{\frac{144}{125}}$$

Divide \frac{1152}{25} by \frac{144}{125} by multiplying \frac{1152}{25} by the reciprocal of \frac{144}{125}.

$$x^{2}+40x=\frac{\frac{1296}{5}}{\frac{144}{125}}$$

Divide \frac{1296}{5} by \frac{144}{125} by multiplying \frac{1296}{5} by the reciprocal of \frac{144}{125}.

$$x^{2}+40x=225$$

Divide 40, the coefficient of the x term, by 2 to get 20. Then add the square of 20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+40x+20^{2}=225+20^{2}$$

Square 20.

$$x^{2}+40x+400=225+400$$

Add 225 to 400.

$$x^{2}+40x+400=625$$

Factor x^{2}+40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+20\right)^{2}=625$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+20\right)^{2}}=\sqrt{625}$$

Simplify.

$$x+20=25 x+20=-25$$

Subtract 20 from both sides of the equation.

$$x=5 x=-45$$