Solve 40=28t+1/210t^2

$$40 = 28 t + \frac { 1 } { 2 } 10 t ^ { 2 }$$

Solve for t

$$t = \frac{6 \sqrt{11} - 14}{5} \approx 1.179949748$$

$$t=\frac{-6\sqrt{11}-14}{5}\approx -6.779949748$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 10 to get 5.

$$40=28t+5t^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$28t+5t^{2}=40$$

Subtract 40 from both sides.

$$28t+5t^{2}-40=0$$

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.

$$5t^{2}+28t-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 28 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-28±\sqrt{28^{2}-4\times 5\left(-40\right)}}{2\times 5}$$

Square 28.

$$t=\frac{-28±\sqrt{784-4\times 5\left(-40\right)}}{2\times 5}$$

Multiply -4 times 5.

$$t=\frac{-28±\sqrt{784-20\left(-40\right)}}{2\times 5}$$

Multiply -20 times -40.

$$t=\frac{-28±\sqrt{784+800}}{2\times 5}$$

Add 784 to 800.

$$t=\frac{-28±\sqrt{1584}}{2\times 5}$$

Take the square root of 1584.

$$t=\frac{-28±12\sqrt{11}}{2\times 5}$$

Multiply 2 times 5.

$$t=\frac{-28±12\sqrt{11}}{10}$$

Now solve the equation t=\frac{-28±12\sqrt{11}}{10} when ± is plus. Add -28 to 12\sqrt{11}.

$$t=\frac{12\sqrt{11}-28}{10}$$

Divide -28+12\sqrt{11} by 10.

$$t=\frac{6\sqrt{11}-14}{5}$$

Now solve the equation t=\frac{-28±12\sqrt{11}}{10} when ± is minus. Subtract 12\sqrt{11} from -28.

$$t=\frac{-12\sqrt{11}-28}{10}$$

Divide -28-12\sqrt{11} by 10.

$$t=\frac{-6\sqrt{11}-14}{5}$$

The equation is now solved.

$$t=\frac{6\sqrt{11}-14}{5} t=\frac{-6\sqrt{11}-14}{5}$$

Steps for Completing the Square

Multiply \frac{1}{2} and 10 to get 5.

$$40=28t+5t^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$28t+5t^{2}=40$$

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.

$$5t^{2}+28t=40$$

Divide both sides by 5.

$$\frac{5t^{2}+28t}{5}=\frac{40}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$t^{2}+\frac{28}{5}t=\frac{40}{5}$$

Divide 40 by 5.

$$t^{2}+\frac{28}{5}t=8$$

Divide \frac{28}{5}, the coefficient of the x term, by 2 to get \frac{14}{5}. Then add the square of \frac{14}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{28}{5}t+\left(\frac{14}{5}\right)^{2}=8+\left(\frac{14}{5}\right)^{2}$$

Square \frac{14}{5} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{28}{5}t+\frac{196}{25}=8+\frac{196}{25}$$

Add 8 to \frac{196}{25}.

$$t^{2}+\frac{28}{5}t+\frac{196}{25}=\frac{396}{25}$$

Factor t^{2}+\frac{28}{5}t+\frac{196}{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}.

$$\left(t+\frac{14}{5}\right)^{2}=\frac{396}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{14}{5}\right)^{2}}=\sqrt{\frac{396}{25}}$$

Simplify.

$$t+\frac{14}{5}=\frac{6\sqrt{11}}{5} t+\frac{14}{5}=-\frac{6\sqrt{11}}{5}$$

Subtract \frac{14}{5} from both sides of the equation.

$$t=\frac{6\sqrt{11}-14}{5} t=\frac{-6\sqrt{11}-14}{5}$$