Solve t+19=t(t+8)

$$t + 19 = t ( t + 8 )$$

Solve for t

$$t = \frac{5 \sqrt{5} - 7}{2} \approx 2.090169944$$

$$t=\frac{-5\sqrt{5}-7}{2}\approx -9.090169944$$

Steps Using the Quadratic Formula

Use the distributive property to multiply t by t+8.

$$t+19=t^{2}+8t$$

Subtract t^{2} from both sides.

$$t+19-t^{2}=8t$$

Subtract 8t from both sides.

$$t+19-t^{2}-8t=0$$

Combine t and -8t to get -7t.

$$-7t+19-t^{2}=0$$

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.

$$-t^{2}-7t+19=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -7 for b, and 19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\times 19}}{2\left(-1\right)}$$

Square -7.

$$t=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\times 19}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{-\left(-7\right)±\sqrt{49+4\times 19}}{2\left(-1\right)}$$

Multiply 4 times 19.

$$t=\frac{-\left(-7\right)±\sqrt{49+76}}{2\left(-1\right)}$$

Add 49 to 76.

$$t=\frac{-\left(-7\right)±\sqrt{125}}{2\left(-1\right)}$$

Take the square root of 125.

$$t=\frac{-\left(-7\right)±5\sqrt{5}}{2\left(-1\right)}$$

The opposite of -7 is 7.

$$t=\frac{7±5\sqrt{5}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{7±5\sqrt{5}}{-2}$$

Now solve the equation t=\frac{7±5\sqrt{5}}{-2} when ± is plus. Add 7 to 5\sqrt{5}.

$$t=\frac{5\sqrt{5}+7}{-2}$$

Divide 7+5\sqrt{5} by -2.

$$t=\frac{-5\sqrt{5}-7}{2}$$

Now solve the equation t=\frac{7±5\sqrt{5}}{-2} when ± is minus. Subtract 5\sqrt{5} from 7.

$$t=\frac{7-5\sqrt{5}}{-2}$$

Divide 7-5\sqrt{5} by -2.

$$t=\frac{5\sqrt{5}-7}{2}$$

The equation is now solved.

$$t=\frac{-5\sqrt{5}-7}{2} t=\frac{5\sqrt{5}-7}{2}$$

Steps for Completing the Square

Use the distributive property to multiply t by t+8.

$$t+19=t^{2}+8t$$

Subtract t^{2} from both sides.

$$t+19-t^{2}=8t$$

Subtract 8t from both sides.

$$t+19-t^{2}-8t=0$$

Combine t and -8t to get -7t.

$$-7t+19-t^{2}=0$$

Subtract 19 from both sides. Anything subtracted from zero gives its negation.

$$-7t-t^{2}=-19$$

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.

$$-t^{2}-7t=-19$$

Divide both sides by -1.

$$\frac{-t^{2}-7t}{-1}=-\frac{19}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$t^{2}+\left(-\frac{7}{-1}\right)t=-\frac{19}{-1}$$

Divide -7 by -1.

$$t^{2}+7t=-\frac{19}{-1}$$

Divide -19 by -1.

$$t^{2}+7t=19$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+7t+\left(\frac{7}{2}\right)^{2}=19+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+7t+\frac{49}{4}=19+\frac{49}{4}$$

Add 19 to \frac{49}{4}.

$$t^{2}+7t+\frac{49}{4}=\frac{125}{4}$$

Factor t^{2}+7t+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{7}{2}\right)^{2}=\frac{125}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{7}{2}\right)^{2}}=\sqrt{\frac{125}{4}}$$

Simplify.

$$t+\frac{7}{2}=\frac{5\sqrt{5}}{2} t+\frac{7}{2}=-\frac{5\sqrt{5}}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$t=\frac{5\sqrt{5}-7}{2} t=\frac{-5\sqrt{5}-7}{2}$$