Solve t/10=4.8-t^2/2/8

$$\frac { t } { 10 } = \frac { \frac { 4.8 - t ^ { 2 } } { 2 } } { 8 }$$

Solve for t

$$t = \frac{2 \sqrt{34} - 4}{5} \approx 1.532380758$$

$$t=\frac{-2\sqrt{34}-4}{5}\approx -3.132380758$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 40, the least common multiple of 10,8.

$$4t=5\times \frac{4.8-t^{2}}{2}$$

Divide each term of 4.8-t^{2} by 2 to get 2.4-\frac{1}{2}t^{2}.

$$4t=5\left(2.4-\frac{1}{2}t^{2}\right)$$

Use the distributive property to multiply 5 by 2.4-\frac{1}{2}t^{2}.

$$4t=12-\frac{5}{2}t^{2}$$

Subtract 12 from both sides.

$$4t-12=-\frac{5}{2}t^{2}$$

Add \frac{5}{2}t^{2} to both sides.

$$4t-12+\frac{5}{2}t^{2}=0$$

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.

$$\frac{5}{2}t^{2}+4t-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{2} for a, 4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-4±\sqrt{4^{2}-4\times \frac{5}{2}\left(-12\right)}}{2\times \frac{5}{2}}$$

Square 4.

$$t=\frac{-4±\sqrt{16-4\times \frac{5}{2}\left(-12\right)}}{2\times \frac{5}{2}}$$

Multiply -4 times \frac{5}{2}.

$$t=\frac{-4±\sqrt{16-10\left(-12\right)}}{2\times \frac{5}{2}}$$

Multiply -10 times -12.

$$t=\frac{-4±\sqrt{16+120}}{2\times \frac{5}{2}}$$

Add 16 to 120.

$$t=\frac{-4±\sqrt{136}}{2\times \frac{5}{2}}$$

Take the square root of 136.

$$t=\frac{-4±2\sqrt{34}}{2\times \frac{5}{2}}$$

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

$$t=\frac{-4±2\sqrt{34}}{5}$$

Now solve the equation t=\frac{-4±2\sqrt{34}}{5} when ± is plus. Add -4 to 2\sqrt{34}.

$$t=\frac{2\sqrt{34}-4}{5}$$

Now solve the equation t=\frac{-4±2\sqrt{34}}{5} when ± is minus. Subtract 2\sqrt{34} from -4.

$$t=\frac{-2\sqrt{34}-4}{5}$$

The equation is now solved.

$$t=\frac{2\sqrt{34}-4}{5} t=\frac{-2\sqrt{34}-4}{5}$$

Steps for Completing the Square

Multiply both sides of the equation by 40, the least common multiple of 10,8.

$$4t=5\times \frac{4.8-t^{2}}{2}$$

Divide each term of 4.8-t^{2} by 2 to get 2.4-\frac{1}{2}t^{2}.

$$4t=5\left(2.4-\frac{1}{2}t^{2}\right)$$

Use the distributive property to multiply 5 by 2.4-\frac{1}{2}t^{2}.

$$4t=12-\frac{5}{2}t^{2}$$

Add \frac{5}{2}t^{2} to both sides.

$$4t+\frac{5}{2}t^{2}=12$$

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.

$$\frac{5}{2}t^{2}+4t=12$$

Divide both sides of the equation by \frac{5}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{2}t^{2}+4t}{\frac{5}{2}}=\frac{12}{\frac{5}{2}}$$

Dividing by \frac{5}{2} undoes the multiplication by \frac{5}{2}.

$$t^{2}+\frac{4}{\frac{5}{2}}t=\frac{12}{\frac{5}{2}}$$

Divide 4 by \frac{5}{2} by multiplying 4 by the reciprocal of \frac{5}{2}.

$$t^{2}+\frac{8}{5}t=\frac{12}{\frac{5}{2}}$$

Divide 12 by \frac{5}{2} by multiplying 12 by the reciprocal of \frac{5}{2}.

$$t^{2}+\frac{8}{5}t=\frac{24}{5}$$

Divide \frac{8}{5}, the coefficient of the x term, by 2 to get \frac{4}{5}. Then add the square of \frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{8}{5}t+\left(\frac{4}{5}\right)^{2}=\frac{24}{5}+\left(\frac{4}{5}\right)^{2}$$

Square \frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{8}{5}t+\frac{16}{25}=\frac{24}{5}+\frac{16}{25}$$

Add \frac{24}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{8}{5}t+\frac{16}{25}=\frac{136}{25}$$

Factor t^{2}+\frac{8}{5}t+\frac{16}{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}.

$$\left(t+\frac{4}{5}\right)^{2}=\frac{136}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{4}{5}\right)^{2}}=\sqrt{\frac{136}{25}}$$

Simplify.

$$t+\frac{4}{5}=\frac{2\sqrt{34}}{5} t+\frac{4}{5}=-\frac{2\sqrt{34}}{5}$$

Subtract \frac{4}{5} from both sides of the equation.

$$t=\frac{2\sqrt{34}-4}{5} t=\frac{-2\sqrt{34}-4}{5}$$