Solve 20t^2-17t=63

$$20 t ^ { 2 } - 17 t = 63$$

Solve for t

$$t = -\frac{7}{5} = -1\frac{2}{5} = -1.4$$

$$t = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

Steps Using Factoring By Grouping

Subtract 63 from both sides.

$$20t^{2}-17t-63=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20t^{2}+at+bt-63. To find a and b, set up a system to be solved.

$$a+b=-17 ab=20\left(-63\right)=-1260$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -1260.

$$1,-1260 2,-630 3,-420 4,-315 5,-252 6,-210 7,-180 9,-140 10,-126 12,-105 14,-90 15,-84 18,-70 20,-63 21,-60 28,-45 30,-42 35,-36$$

Calculate the sum for each pair.

$$1-1260=-1259 2-630=-628 3-420=-417 4-315=-311 5-252=-247 6-210=-204 7-180=-173 9-140=-131 10-126=-116 12-105=-93 14-90=-76 15-84=-69 18-70=-52 20-63=-43 21-60=-39 28-45=-17 30-42=-12 35-36=-1$$

The solution is the pair that gives sum -17.

$$a=-45 b=28$$

Rewrite 20t^{2}-17t-63 as \left(20t^{2}-45t\right)+\left(28t-63\right).

$$\left(20t^{2}-45t\right)+\left(28t-63\right)$$

Factor out 5t in the first and 7 in the second group.

$$5t\left(4t-9\right)+7\left(4t-9\right)$$

Factor out common term 4t-9 by using distributive property.

$$\left(4t-9\right)\left(5t+7\right)$$

To find equation solutions, solve 4t-9=0 and 5t+7=0.

$$t=\frac{9}{4} t=-\frac{7}{5}$$

Steps Using the Quadratic Formula

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.

$$20t^{2}-17t=63$$

Subtract 63 from both sides of the equation.

$$20t^{2}-17t-63=63-63$$

Subtracting 63 from itself leaves 0.

$$20t^{2}-17t-63=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -17 for b, and -63 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 20\left(-63\right)}}{2\times 20}$$

Square -17.

$$t=\frac{-\left(-17\right)±\sqrt{289-4\times 20\left(-63\right)}}{2\times 20}$$

Multiply -4 times 20.

$$t=\frac{-\left(-17\right)±\sqrt{289-80\left(-63\right)}}{2\times 20}$$

Multiply -80 times -63.

$$t=\frac{-\left(-17\right)±\sqrt{289+5040}}{2\times 20}$$

Add 289 to 5040.

$$t=\frac{-\left(-17\right)±\sqrt{5329}}{2\times 20}$$

Take the square root of 5329.

$$t=\frac{-\left(-17\right)±73}{2\times 20}$$

The opposite of -17 is 17.

$$t=\frac{17±73}{2\times 20}$$

Multiply 2 times 20.

$$t=\frac{17±73}{40}$$

Now solve the equation t=\frac{17±73}{40} when ± is plus. Add 17 to 73.

$$t=\frac{90}{40}$$

Reduce the fraction \frac{90}{40} to lowest terms by extracting and canceling out 10.

$$t=\frac{9}{4}$$

Now solve the equation t=\frac{17±73}{40} when ± is minus. Subtract 73 from 17.

$$t=-\frac{56}{40}$$

Reduce the fraction \frac{-56}{40} to lowest terms by extracting and canceling out 8.

$$t=-\frac{7}{5}$$

The equation is now solved.

$$t=\frac{9}{4} t=-\frac{7}{5}$$

Steps for Completing the Square

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.

$$20t^{2}-17t=63$$

Divide both sides by 20.

$$\frac{20t^{2}-17t}{20}=\frac{63}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$t^{2}-\frac{17}{20}t=\frac{63}{20}$$

Divide -\frac{17}{20}, the coefficient of the x term, by 2 to get -\frac{17}{40}. Then add the square of -\frac{17}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{17}{20}t+\left(-\frac{17}{40}\right)^{2}=\frac{63}{20}+\left(-\frac{17}{40}\right)^{2}$$

Square -\frac{17}{40} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{17}{20}t+\frac{289}{1600}=\frac{63}{20}+\frac{289}{1600}$$

Add \frac{63}{20} to \frac{289}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{17}{20}t+\frac{289}{1600}=\frac{5329}{1600}$$

Factor t^{2}-\frac{17}{20}t+\frac{289}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{17}{40}\right)^{2}=\frac{5329}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{17}{40}\right)^{2}}=\sqrt{\frac{5329}{1600}}$$

Simplify.

$$t-\frac{17}{40}=\frac{73}{40} t-\frac{17}{40}=-\frac{73}{40}$$

Add \frac{17}{40} to both sides of the equation.

$$t=\frac{9}{4} t=-\frac{7}{5}$$