Solve t+9=t(t+9)

$$t + 9 = t ( t + 9 )$$

Solve for t

$$t=-9$$

$$t=1$$

Steps Using Factoring By Grouping

Use the distributive property to multiply t by t+9.

$$t+9=t^{2}+9t$$

Subtract t^{2} from both sides.

$$t+9-t^{2}=9t$$

Subtract 9t from both sides.

$$t+9-t^{2}-9t=0$$

Combine t and -9t to get -8t.

$$-8t+9-t^{2}=0$$

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

$$-t^{2}-8t+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -t^{2}+at+bt+9. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-9=-9$$

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 -9.

$$1,-9 3,-3$$

Calculate the sum for each pair.

$$1-9=-8 3-3=0$$

The solution is the pair that gives sum -8.

$$a=1 b=-9$$

Rewrite -t^{2}-8t+9 as \left(-t^{2}+t\right)+\left(-9t+9\right).

$$\left(-t^{2}+t\right)+\left(-9t+9\right)$$

Factor out t in the first and 9 in the second group.

$$t\left(-t+1\right)+9\left(-t+1\right)$$

Factor out common term -t+1 by using distributive property.

$$\left(-t+1\right)\left(t+9\right)$$

To find equation solutions, solve -t+1=0 and t+9=0.

$$t=1 t=-9$$

Steps Using the Quadratic Formula

Use the distributive property to multiply t by t+9.

$$t+9=t^{2}+9t$$

Subtract t^{2} from both sides.

$$t+9-t^{2}=9t$$

Subtract 9t from both sides.

$$t+9-t^{2}-9t=0$$

Combine t and -9t to get -8t.

$$-8t+9-t^{2}=0$$

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.

$$-t^{2}-8t+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -8 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Square -8.

$$t=\frac{-\left(-8\right)±\sqrt{64-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-\left(-8\right)±\sqrt{64+4\times 9}}{2\left(-1\right)}$$

Multiply 4 times 9.

$$t=\frac{-\left(-8\right)±\sqrt{64+36}}{2\left(-1\right)}$$

Add 64 to 36.

$$t=\frac{-\left(-8\right)±\sqrt{100}}{2\left(-1\right)}$$

Take the square root of 100.

$$t=\frac{-\left(-8\right)±10}{2\left(-1\right)}$$

The opposite of -8 is 8.

$$t=\frac{8±10}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{8±10}{-2}$$

Now solve the equation t=\frac{8±10}{-2} when ± is plus. Add 8 to 10.

$$t=\frac{18}{-2}$$

Divide 18 by -2.

$$t=-9$$

Now solve the equation t=\frac{8±10}{-2} when ± is minus. Subtract 10 from 8.

$$t=-\frac{2}{-2}$$

Divide -2 by -2.

$$t=1$$

The equation is now solved.

$$t=-9 t=1$$

Steps for Completing the Square

Use the distributive property to multiply t by t+9.

$$t+9=t^{2}+9t$$

Subtract t^{2} from both sides.

$$t+9-t^{2}=9t$$

Subtract 9t from both sides.

$$t+9-t^{2}-9t=0$$

Combine t and -9t to get -8t.

$$-8t+9-t^{2}=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-8t-t^{2}=-9$$

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.

$$-t^{2}-8t=-9$$

Divide both sides by -1.

$$\frac{-t^{2}-8t}{-1}=-\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\left(-\frac{8}{-1}\right)t=-\frac{9}{-1}$$

Divide -8 by -1.

$$t^{2}+8t=-\frac{9}{-1}$$

Divide -9 by -1.

$$t^{2}+8t=9$$

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

$$t^{2}+8t+4^{2}=9+4^{2}$$

Square 4.

$$t^{2}+8t+16=9+16$$

Add 9 to 16.

$$t^{2}+8t+16=25$$

Factor t^{2}+8t+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$t+4=5 t+4=-5$$

Subtract 4 from both sides of the equation.

$$t=1 t=-9$$