Solve 4t^2-128/5t+64=9/5t^2

$$4 t ^ { 2 } - \frac { 128 } { 5 } t + 64 = \frac { 9 } { 5 } t ^ { 2 }$$

Solve for t

$$t=8$$

$$t = \frac{40}{11} = 3\frac{7}{11} \approx 3.636363636$$

Steps Using the Quadratic Formula

Subtract \frac{9}{5}t^{2} from both sides.

$$4t^{2}-\frac{128}{5}t+64-\frac{9}{5}t^{2}=0$$

Combine 4t^{2} and -\frac{9}{5}t^{2} to get \frac{11}{5}t^{2}.

$$\frac{11}{5}t^{2}-\frac{128}{5}t+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{11}{5} for a, -\frac{128}{5} for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\left(-\frac{128}{5}\right)^{2}-4\times \frac{11}{5}\times 64}}{2\times \frac{11}{5}}$$

Square -\frac{128}{5} by squaring both the numerator and the denominator of the fraction.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\frac{16384}{25}-4\times \frac{11}{5}\times 64}}{2\times \frac{11}{5}}$$

Multiply -4 times \frac{11}{5}.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\frac{16384}{25}-\frac{44}{5}\times 64}}{2\times \frac{11}{5}}$$

Multiply -\frac{44}{5} times 64.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\frac{16384}{25}-\frac{2816}{5}}}{2\times \frac{11}{5}}$$

Add \frac{16384}{25} to -\frac{2816}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-\left(-\frac{128}{5}\right)±\sqrt{\frac{2304}{25}}}{2\times \frac{11}{5}}$$

Take the square root of \frac{2304}{25}.

$$t=\frac{-\left(-\frac{128}{5}\right)±\frac{48}{5}}{2\times \frac{11}{5}}$$

The opposite of -\frac{128}{5} is \frac{128}{5}.

$$t=\frac{\frac{128}{5}±\frac{48}{5}}{2\times \frac{11}{5}}$$

Multiply 2 times \frac{11}{5}.

$$t=\frac{\frac{128}{5}±\frac{48}{5}}{\frac{22}{5}}$$

Now solve the equation t=\frac{\frac{128}{5}±\frac{48}{5}}{\frac{22}{5}} when ± is plus. Add \frac{128}{5} to \frac{48}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{\frac{176}{5}}{\frac{22}{5}}$$

Divide \frac{176}{5} by \frac{22}{5} by multiplying \frac{176}{5} by the reciprocal of \frac{22}{5}.

$$t=8$$

Now solve the equation t=\frac{\frac{128}{5}±\frac{48}{5}}{\frac{22}{5}} when ± is minus. Subtract \frac{48}{5} from \frac{128}{5} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{16}{\frac{22}{5}}$$

Divide 16 by \frac{22}{5} by multiplying 16 by the reciprocal of \frac{22}{5}.

$$t=\frac{40}{11}$$

The equation is now solved.

$$t=8 t=\frac{40}{11}$$

Steps for Completing the Square

Subtract \frac{9}{5}t^{2} from both sides.

$$4t^{2}-\frac{128}{5}t+64-\frac{9}{5}t^{2}=0$$

Combine 4t^{2} and -\frac{9}{5}t^{2} to get \frac{11}{5}t^{2}.

$$\frac{11}{5}t^{2}-\frac{128}{5}t+64=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$\frac{11}{5}t^{2}-\frac{128}{5}t=-64$$

Divide both sides of the equation by \frac{11}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{11}{5}t^{2}-\frac{128}{5}t}{\frac{11}{5}}=-\frac{64}{\frac{11}{5}}$$

Dividing by \frac{11}{5} undoes the multiplication by \frac{11}{5}.

$$t^{2}+\left(-\frac{\frac{128}{5}}{\frac{11}{5}}\right)t=-\frac{64}{\frac{11}{5}}$$

Divide -\frac{128}{5} by \frac{11}{5} by multiplying -\frac{128}{5} by the reciprocal of \frac{11}{5}.

$$t^{2}-\frac{128}{11}t=-\frac{64}{\frac{11}{5}}$$

Divide -64 by \frac{11}{5} by multiplying -64 by the reciprocal of \frac{11}{5}.

$$t^{2}-\frac{128}{11}t=-\frac{320}{11}$$

Divide -\frac{128}{11}, the coefficient of the x term, by 2 to get -\frac{64}{11}. Then add the square of -\frac{64}{11} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{128}{11}t+\left(-\frac{64}{11}\right)^{2}=-\frac{320}{11}+\left(-\frac{64}{11}\right)^{2}$$

Square -\frac{64}{11} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{128}{11}t+\frac{4096}{121}=-\frac{320}{11}+\frac{4096}{121}$$

Add -\frac{320}{11} to \frac{4096}{121} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{128}{11}t+\frac{4096}{121}=\frac{576}{121}$$

Factor t^{2}-\frac{128}{11}t+\frac{4096}{121}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{64}{11}\right)^{2}=\frac{576}{121}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{64}{11}\right)^{2}}=\sqrt{\frac{576}{121}}$$

Simplify.

$$t-\frac{64}{11}=\frac{24}{11} t-\frac{64}{11}=-\frac{24}{11}$$

Add \frac{64}{11} to both sides of the equation.

$$t=8 t=\frac{40}{11}$$