Solve 17=22t-5t^2

$$17 = 22 t - 5 t ^ { 2 }$$

Solve for t

$$t=1$$

$$t = \frac{17}{5} = 3\frac{2}{5} = 3.4$$

Steps Using Factoring By Grouping

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

$$22t-5t^{2}=17$$

Subtract 17 from both sides.

$$22t-5t^{2}-17=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-5t^{2}+22t-17=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5t^{2}+at+bt-17. To find a and b, set up a system to be solved.

$$a+b=22 ab=-5\left(-17\right)=85$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 85.

$$1,85 5,17$$

Calculate the sum for each pair.

$$1+85=86 5+17=22$$

The solution is the pair that gives sum 22.

$$a=17 b=5$$

Rewrite -5t^{2}+22t-17 as \left(-5t^{2}+17t\right)+\left(5t-17\right).

$$\left(-5t^{2}+17t\right)+\left(5t-17\right)$$

Factor out -t in -5t^{2}+17t.

$$-t\left(5t-17\right)+5t-17$$

Factor out common term 5t-17 by using distributive property.

$$\left(5t-17\right)\left(-t+1\right)$$

To find equation solutions, solve 5t-17=0 and -t+1=0.

$$t=\frac{17}{5} t=1$$

Steps Using the Quadratic Formula

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

$$22t-5t^{2}=17$$

Subtract 17 from both sides.

$$22t-5t^{2}-17=0$$

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.

$$-5t^{2}+22t-17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 22 for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-22±\sqrt{22^{2}-4\left(-5\right)\left(-17\right)}}{2\left(-5\right)}$$

Square 22.

$$t=\frac{-22±\sqrt{484-4\left(-5\right)\left(-17\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-22±\sqrt{484+20\left(-17\right)}}{2\left(-5\right)}$$

Multiply 20 times -17.

$$t=\frac{-22±\sqrt{484-340}}{2\left(-5\right)}$$

Add 484 to -340.

$$t=\frac{-22±\sqrt{144}}{2\left(-5\right)}$$

Take the square root of 144.

$$t=\frac{-22±12}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-22±12}{-10}$$

Now solve the equation t=\frac{-22±12}{-10} when ± is plus. Add -22 to 12.

$$t=-\frac{10}{-10}$$

Divide -10 by -10.

$$t=1$$

Now solve the equation t=\frac{-22±12}{-10} when ± is minus. Subtract 12 from -22.

$$t=-\frac{34}{-10}$$

Reduce the fraction \frac{-34}{-10} to lowest terms by extracting and canceling out 2.

$$t=\frac{17}{5}$$

The equation is now solved.

$$t=1 t=\frac{17}{5}$$

Steps for Completing the Square

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

$$22t-5t^{2}=17$$

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.

$$-5t^{2}+22t=17$$

Divide both sides by -5.

$$\frac{-5t^{2}+22t}{-5}=\frac{17}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\frac{22}{-5}t=\frac{17}{-5}$$

Divide 22 by -5.

$$t^{2}-\frac{22}{5}t=\frac{17}{-5}$$

Divide 17 by -5.

$$t^{2}-\frac{22}{5}t=-\frac{17}{5}$$

Divide -\frac{22}{5}, the coefficient of the x term, by 2 to get -\frac{11}{5}. Then add the square of -\frac{11}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{22}{5}t+\left(-\frac{11}{5}\right)^{2}=-\frac{17}{5}+\left(-\frac{11}{5}\right)^{2}$$

Square -\frac{11}{5} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{22}{5}t+\frac{121}{25}=-\frac{17}{5}+\frac{121}{25}$$

Add -\frac{17}{5} to \frac{121}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{22}{5}t+\frac{121}{25}=\frac{36}{25}$$

Factor t^{2}-\frac{22}{5}t+\frac{121}{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}.

$$\left(t-\frac{11}{5}\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{11}{5}\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

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

Add \frac{11}{5} to both sides of the equation.

$$t=\frac{17}{5} t=1$$