Solve 0.6t-160/3*5*10^-4/4*10^-3t^2=-2.04

$$0.6 t - \frac { \frac { 160 } { 3 } \times 5 \times 10 ^ { - 4 } } { 4 \times 10 ^ { - 3 } } t ^ { 2 } = - 2.04$$

Solve for t

$$t=-0.51$$

$$t=0.6$$

Steps Using the Quadratic Formula

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$0.6t-\frac{5\times \frac{160}{3}}{4\times 10^{1}}t^{2}=-2.04$$

Multiply 5 and \frac{160}{3} to get \frac{800}{3}.

$$0.6t-\frac{\frac{800}{3}}{4\times 10^{1}}t^{2}=-2.04$$

Calculate 10 to the power of 1 and get 10.

$$0.6t-\frac{\frac{800}{3}}{4\times 10}t^{2}=-2.04$$

Multiply 4 and 10 to get 40.

$$0.6t-\frac{\frac{800}{3}}{40}t^{2}=-2.04$$

Express \frac{\frac{800}{3}}{40} as a single fraction.

$$0.6t-\frac{800}{3\times 40}t^{2}=-2.04$$

Multiply 3 and 40 to get 120.

$$0.6t-\frac{800}{120}t^{2}=-2.04$$

Reduce the fraction \frac{800}{120} to lowest terms by extracting and canceling out 40.

$$0.6t-\frac{20}{3}t^{2}=-2.04$$

Add 2.04 to both sides.

$$0.6t-\frac{20}{3}t^{2}+2.04=0$$

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.

$$-\frac{20}{3}t^{2}+\frac{3}{5}t+2.04=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{20}{3} for a, \frac{3}{5} for b, and 2.04 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\frac{3}{5}±\sqrt{\left(\frac{3}{5}\right)^{2}-4\left(-\frac{20}{3}\right)\times 2.04}}{2\left(-\frac{20}{3}\right)}$$

Square \frac{3}{5} by squaring both the numerator and the denominator of the fraction.

$$t=\frac{-\frac{3}{5}±\sqrt{\frac{9}{25}-4\left(-\frac{20}{3}\right)\times 2.04}}{2\left(-\frac{20}{3}\right)}$$

Multiply -4 times -\frac{20}{3}.

$$t=\frac{-\frac{3}{5}±\sqrt{\frac{9}{25}+\frac{80}{3}\times 2.04}}{2\left(-\frac{20}{3}\right)}$$

Multiply \frac{80}{3} times 2.04 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-\frac{3}{5}±\sqrt{\frac{9}{25}+\frac{272}{5}}}{2\left(-\frac{20}{3}\right)}$$

Add \frac{9}{25} to \frac{272}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{-\frac{3}{5}±\sqrt{\frac{1369}{25}}}{2\left(-\frac{20}{3}\right)}$$

Take the square root of \frac{1369}{25}.

$$t=\frac{-\frac{3}{5}±\frac{37}{5}}{2\left(-\frac{20}{3}\right)}$$

Multiply 2 times -\frac{20}{3}.

$$t=\frac{-\frac{3}{5}±\frac{37}{5}}{-\frac{40}{3}}$$

Now solve the equation t=\frac{-\frac{3}{5}±\frac{37}{5}}{-\frac{40}{3}} when ± is plus. Add -\frac{3}{5} to \frac{37}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t=\frac{\frac{34}{5}}{-\frac{40}{3}}$$

Divide \frac{34}{5} by -\frac{40}{3} by multiplying \frac{34}{5} by the reciprocal of -\frac{40}{3}.

$$t=-\frac{51}{100}$$

Now solve the equation t=\frac{-\frac{3}{5}±\frac{37}{5}}{-\frac{40}{3}} when ± is minus. Subtract \frac{37}{5} from -\frac{3}{5} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$t=-\frac{8}{-\frac{40}{3}}$$

Divide -8 by -\frac{40}{3} by multiplying -8 by the reciprocal of -\frac{40}{3}.

$$t=\frac{3}{5}$$

The equation is now solved.

$$t=-\frac{51}{100} t=\frac{3}{5}$$

Steps for Completing the Square

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$0.6t-\frac{5\times \frac{160}{3}}{4\times 10^{1}}t^{2}=-2.04$$

Multiply 5 and \frac{160}{3} to get \frac{800}{3}.

$$0.6t-\frac{\frac{800}{3}}{4\times 10^{1}}t^{2}=-2.04$$

Calculate 10 to the power of 1 and get 10.

$$0.6t-\frac{\frac{800}{3}}{4\times 10}t^{2}=-2.04$$

Multiply 4 and 10 to get 40.

$$0.6t-\frac{\frac{800}{3}}{40}t^{2}=-2.04$$

Express \frac{\frac{800}{3}}{40} as a single fraction.

$$0.6t-\frac{800}{3\times 40}t^{2}=-2.04$$

Multiply 3 and 40 to get 120.

$$0.6t-\frac{800}{120}t^{2}=-2.04$$

Reduce the fraction \frac{800}{120} to lowest terms by extracting and canceling out 40.

$$0.6t-\frac{20}{3}t^{2}=-2.04$$

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.

$$-\frac{20}{3}t^{2}+\frac{3}{5}t=-2.04$$

Divide both sides of the equation by -\frac{20}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{20}{3}t^{2}+\frac{3}{5}t}{-\frac{20}{3}}=-\frac{2.04}{-\frac{20}{3}}$$

Dividing by -\frac{20}{3} undoes the multiplication by -\frac{20}{3}.

$$t^{2}+\frac{\frac{3}{5}}{-\frac{20}{3}}t=-\frac{2.04}{-\frac{20}{3}}$$

Divide \frac{3}{5} by -\frac{20}{3} by multiplying \frac{3}{5} by the reciprocal of -\frac{20}{3}.

$$t^{2}-\frac{9}{100}t=-\frac{2.04}{-\frac{20}{3}}$$

Divide -2.04 by -\frac{20}{3} by multiplying -2.04 by the reciprocal of -\frac{20}{3}.

$$t^{2}-\frac{9}{100}t=\frac{153}{500}$$

Divide -\frac{9}{100}, the coefficient of the x term, by 2 to get -\frac{9}{200}. Then add the square of -\frac{9}{200} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{9}{100}t+\left(-\frac{9}{200}\right)^{2}=\frac{153}{500}+\left(-\frac{9}{200}\right)^{2}$$

Square -\frac{9}{200} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{9}{100}t+\frac{81}{40000}=\frac{153}{500}+\frac{81}{40000}$$

Add \frac{153}{500} to \frac{81}{40000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{9}{100}t+\frac{81}{40000}=\frac{12321}{40000}$$

Factor t^{2}-\frac{9}{100}t+\frac{81}{40000}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{9}{200}\right)^{2}=\frac{12321}{40000}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{9}{200}\right)^{2}}=\sqrt{\frac{12321}{40000}}$$

Simplify.

$$t-\frac{9}{200}=\frac{111}{200} t-\frac{9}{200}=-\frac{111}{200}$$

Add \frac{9}{200} to both sides of the equation.

$$t=\frac{3}{5} t=-\frac{51}{100}$$