Solve 100=5t+2t^2

$$100 = 5t+2 { t }^{ 2 }$$

Solve for t

$$t = \frac{5 \sqrt{33} - 5}{4} \approx 5.930703308$$

$$t=\frac{-5\sqrt{33}-5}{4}\approx -8.430703308$$

Steps Using the Quadratic Formula

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

$$5t+2t^{2}=100$$

Subtract 100 from both sides.

$$5t+2t^{2}-100=0$$

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.

$$2t^{2}+5t-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 5 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-5±\sqrt{5^{2}-4\times 2\left(-100\right)}}{2\times 2}$$

Square 5.

$$t=\frac{-5±\sqrt{25-4\times 2\left(-100\right)}}{2\times 2}$$

Multiply -4 times 2.

$$t=\frac{-5±\sqrt{25-8\left(-100\right)}}{2\times 2}$$

Multiply -8 times -100.

$$t=\frac{-5±\sqrt{25+800}}{2\times 2}$$

Add 25 to 800.

$$t=\frac{-5±\sqrt{825}}{2\times 2}$$

Take the square root of 825.

$$t=\frac{-5±5\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{-5±5\sqrt{33}}{4}$$

Now solve the equation t=\frac{-5±5\sqrt{33}}{4} when ± is plus. Add -5 to 5\sqrt{33}.

$$t=\frac{5\sqrt{33}-5}{4}$$

Now solve the equation t=\frac{-5±5\sqrt{33}}{4} when ± is minus. Subtract 5\sqrt{33} from -5.

$$t=\frac{-5\sqrt{33}-5}{4}$$

The equation is now solved.

$$t=\frac{5\sqrt{33}-5}{4} t=\frac{-5\sqrt{33}-5}{4}$$

Steps for Completing the Square

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

$$5t+2t^{2}=100$$

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.

$$2t^{2}+5t=100$$

Divide both sides by 2.

$$\frac{2t^{2}+5t}{2}=\frac{100}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$t^{2}+\frac{5}{2}t=\frac{100}{2}$$

Divide 100 by 2.

$$t^{2}+\frac{5}{2}t=50$$

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

$$t^{2}+\frac{5}{2}t+\left(\frac{5}{4}\right)^{2}=50+\left(\frac{5}{4}\right)^{2}$$

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

$$t^{2}+\frac{5}{2}t+\frac{25}{16}=50+\frac{25}{16}$$

Add 50 to \frac{25}{16}.

$$t^{2}+\frac{5}{2}t+\frac{25}{16}=\frac{825}{16}$$

Factor t^{2}+\frac{5}{2}t+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{5}{4}\right)^{2}=\frac{825}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{5}{4}\right)^{2}}=\sqrt{\frac{825}{16}}$$

Simplify.

$$t+\frac{5}{4}=\frac{5\sqrt{33}}{4} t+\frac{5}{4}=-\frac{5\sqrt{33}}{4}$$

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

$$t=\frac{5\sqrt{33}-5}{4} t=\frac{-5\sqrt{33}-5}{4}$$