Solve 17=12t-5t^2

$$17 = 12 t - 5 t ^ { 2 }$$

Solve for t

$$t=\frac{6}{5}+\frac{7}{5}i=1.2+1.4i$$

$$t=\frac{6}{5}-\frac{7}{5}i=1.2-1.4i$$

Steps Using the Quadratic Formula

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

$$12t-5t^{2}=17$$

Subtract 17 from both sides.

$$12t-5t^{2}-17=0$$

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.

$$-5t^{2}+12t-17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 12 for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-12±\sqrt{12^{2}-4\left(-5\right)\left(-17\right)}}{2\left(-5\right)}$$

Square 12.

$$t=\frac{-12±\sqrt{144-4\left(-5\right)\left(-17\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-12±\sqrt{144+20\left(-17\right)}}{2\left(-5\right)}$$

Multiply 20 times -17.

$$t=\frac{-12±\sqrt{144-340}}{2\left(-5\right)}$$

Add 144 to -340.

$$t=\frac{-12±\sqrt{-196}}{2\left(-5\right)}$$

Take the square root of -196.

$$t=\frac{-12±14i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-12±14i}{-10}$$

Now solve the equation t=\frac{-12±14i}{-10} when ± is plus. Add -12 to 14i.

$$t=\frac{-12+14i}{-10}$$

Divide -12+14i by -10.

$$t=\frac{6}{5}-\frac{7}{5}i$$

Now solve the equation t=\frac{-12±14i}{-10} when ± is minus. Subtract 14i from -12.

$$t=\frac{-12-14i}{-10}$$

Divide -12-14i by -10.

$$t=\frac{6}{5}+\frac{7}{5}i$$

The equation is now solved.

$$t=\frac{6}{5}-\frac{7}{5}i t=\frac{6}{5}+\frac{7}{5}i$$

Steps for Completing the Square

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

$$12t-5t^{2}=17$$

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.

$$-5t^{2}+12t=17$$

Divide both sides by -5.

$$\frac{-5t^{2}+12t}{-5}=\frac{17}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\frac{12}{-5}t=\frac{17}{-5}$$

Divide 12 by -5.

$$t^{2}-\frac{12}{5}t=\frac{17}{-5}$$

Divide 17 by -5.

$$t^{2}-\frac{12}{5}t=-\frac{17}{5}$$

Divide -\frac{12}{5}, the coefficient of the x term, by 2 to get -\frac{6}{5}. Then add the square of -\frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{12}{5}t+\left(-\frac{6}{5}\right)^{2}=-\frac{17}{5}+\left(-\frac{6}{5}\right)^{2}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{12}{5}t+\frac{36}{25}=-\frac{17}{5}+\frac{36}{25}$$

Add -\frac{17}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{12}{5}t+\frac{36}{25}=-\frac{49}{25}$$

Factor t^{2}-\frac{12}{5}t+\frac{36}{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}.

$$\left(t-\frac{6}{5}\right)^{2}=-\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{6}{5}\right)^{2}}=\sqrt{-\frac{49}{25}}$$

Simplify.

$$t-\frac{6}{5}=\frac{7}{5}i t-\frac{6}{5}=-\frac{7}{5}i$$

Add \frac{6}{5} to both sides of the equation.

$$t=\frac{6}{5}+\frac{7}{5}i t=\frac{6}{5}-\frac{7}{5}i$$