Solve -(4/3t-1)^2=1/9(28t-35/4)

$$- ( \frac { 4 } { 3 } t - 1 ) ^ { 2 } = \frac { 1 } { 9 } ( 28 t - \frac { 35 } { 4 } )$$

Solve for t

$$t=-\frac{1}{8}=-0.125$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{4}{3}t-1\right)^{2}.

$$-\left(\frac{16}{9}t^{2}-\frac{8}{3}t+1\right)=\frac{1}{9}\left(28t-\frac{35}{4}\right)$$

To find the opposite of \frac{16}{9}t^{2}-\frac{8}{3}t+1, find the opposite of each term.

$$-\frac{16}{9}t^{2}+\frac{8}{3}t-1=\frac{1}{9}\left(28t-\frac{35}{4}\right)$$

Use the distributive property to multiply \frac{1}{9} by 28t-\frac{35}{4}.

$$-\frac{16}{9}t^{2}+\frac{8}{3}t-1=\frac{28}{9}t-\frac{35}{36}$$

Subtract \frac{28}{9}t from both sides.

$$-\frac{16}{9}t^{2}+\frac{8}{3}t-1-\frac{28}{9}t=-\frac{35}{36}$$

Combine \frac{8}{3}t and -\frac{28}{9}t to get -\frac{4}{9}t.

$$-\frac{16}{9}t^{2}-\frac{4}{9}t-1=-\frac{35}{36}$$

Add \frac{35}{36} to both sides.

$$-\frac{16}{9}t^{2}-\frac{4}{9}t-1+\frac{35}{36}=0$$

Add -1 and \frac{35}{36} to get -\frac{1}{36}.

$$-\frac{16}{9}t^{2}-\frac{4}{9}t-\frac{1}{36}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{16}{9} for a, -\frac{4}{9} for b, and -\frac{1}{36} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-\frac{4}{9}\right)±\sqrt{\left(-\frac{4}{9}\right)^{2}-4\left(-\frac{16}{9}\right)\left(-\frac{1}{36}\right)}}{2\left(-\frac{16}{9}\right)}$$

Square -\frac{4}{9} by squaring both the numerator and the denominator of the fraction.

$$t=\frac{-\left(-\frac{4}{9}\right)±\sqrt{\frac{16}{81}-4\left(-\frac{16}{9}\right)\left(-\frac{1}{36}\right)}}{2\left(-\frac{16}{9}\right)}$$

Multiply -4 times -\frac{16}{9}.

$$t=\frac{-\left(-\frac{4}{9}\right)±\sqrt{\frac{16}{81}+\frac{64}{9}\left(-\frac{1}{36}\right)}}{2\left(-\frac{16}{9}\right)}$$

Multiply \frac{64}{9} times -\frac{1}{36} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-\left(-\frac{4}{9}\right)±\sqrt{\frac{16-16}{81}}}{2\left(-\frac{16}{9}\right)}$$

Add \frac{16}{81} to -\frac{16}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-\left(-\frac{4}{9}\right)±\sqrt{0}}{2\left(-\frac{16}{9}\right)}$$

Take the square root of 0.

$$t=-\frac{-\frac{4}{9}}{2\left(-\frac{16}{9}\right)}$$

The opposite of -\frac{4}{9} is \frac{4}{9}.

$$t=\frac{\frac{4}{9}}{2\left(-\frac{16}{9}\right)}$$

Multiply 2 times -\frac{16}{9}.

$$t=\frac{\frac{4}{9}}{-\frac{32}{9}}$$

Divide \frac{4}{9} by -\frac{32}{9} by multiplying \frac{4}{9} by the reciprocal of -\frac{32}{9}.

$$t=-\frac{1}{8}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{4}{3}t-1\right)^{2}.

$$-\left(\frac{16}{9}t^{2}-\frac{8}{3}t+1\right)=\frac{1}{9}\left(28t-\frac{35}{4}\right)$$

To find the opposite of \frac{16}{9}t^{2}-\frac{8}{3}t+1, find the opposite of each term.

$$-\frac{16}{9}t^{2}+\frac{8}{3}t-1=\frac{1}{9}\left(28t-\frac{35}{4}\right)$$

Use the distributive property to multiply \frac{1}{9} by 28t-\frac{35}{4}.

$$-\frac{16}{9}t^{2}+\frac{8}{3}t-1=\frac{28}{9}t-\frac{35}{36}$$

Subtract \frac{28}{9}t from both sides.

$$-\frac{16}{9}t^{2}+\frac{8}{3}t-1-\frac{28}{9}t=-\frac{35}{36}$$

Combine \frac{8}{3}t and -\frac{28}{9}t to get -\frac{4}{9}t.

$$-\frac{16}{9}t^{2}-\frac{4}{9}t-1=-\frac{35}{36}$$

Add 1 to both sides.

$$-\frac{16}{9}t^{2}-\frac{4}{9}t=-\frac{35}{36}+1$$

Add -\frac{35}{36} and 1 to get \frac{1}{36}.

$$-\frac{16}{9}t^{2}-\frac{4}{9}t=\frac{1}{36}$$

Divide both sides of the equation by -\frac{16}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{16}{9}t^{2}-\frac{4}{9}t}{-\frac{16}{9}}=\frac{\frac{1}{36}}{-\frac{16}{9}}$$

Dividing by -\frac{16}{9} undoes the multiplication by -\frac{16}{9}.

$$t^{2}+\left(-\frac{\frac{4}{9}}{-\frac{16}{9}}\right)t=\frac{\frac{1}{36}}{-\frac{16}{9}}$$

Divide -\frac{4}{9} by -\frac{16}{9} by multiplying -\frac{4}{9} by the reciprocal of -\frac{16}{9}.

$$t^{2}+\frac{1}{4}t=\frac{\frac{1}{36}}{-\frac{16}{9}}$$

Divide \frac{1}{36} by -\frac{16}{9} by multiplying \frac{1}{36} by the reciprocal of -\frac{16}{9}.

$$t^{2}+\frac{1}{4}t=-\frac{1}{64}$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{1}{4}t+\left(\frac{1}{8}\right)^{2}=-\frac{1}{64}+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{1}{4}t+\frac{1}{64}=\frac{-1+1}{64}$$

Add -\frac{1}{64} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{1}{4}t+\frac{1}{64}=0$$

Factor t^{2}+\frac{1}{4}t+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{1}{8}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{1}{8}\right)^{2}}=\sqrt{0}$$

Simplify.

$$t+\frac{1}{8}=0 t+\frac{1}{8}=0$$

Subtract \frac{1}{8} from both sides of the equation.

$$t=-\frac{1}{8} t=-\frac{1}{8}$$

The equation is now solved. Solutions are the same.

$$t=-\frac{1}{8}$$