Solve 27=22t-5t^2

$$27 = 22 t - 5 t ^ { 2 }$$

Solve for t

$$t=\frac{11+\sqrt{14}i}{5}\approx 2.2+0.748331477i$$

$$t=\frac{-\sqrt{14}i+11}{5}\approx 2.2-0.748331477i$$

Steps Using the Quadratic Formula

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

$$22t-5t^{2}=27$$

Subtract 27 from both sides.

$$22t-5t^{2}-27=0$$

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.

$$-5t^{2}+22t-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 22 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-22±\sqrt{22^{2}-4\left(-5\right)\left(-27\right)}}{2\left(-5\right)}$$

Square 22.

$$t=\frac{-22±\sqrt{484-4\left(-5\right)\left(-27\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$t=\frac{-22±\sqrt{484+20\left(-27\right)}}{2\left(-5\right)}$$

Multiply 20 times -27.

$$t=\frac{-22±\sqrt{484-540}}{2\left(-5\right)}$$

Add 484 to -540.

$$t=\frac{-22±\sqrt{-56}}{2\left(-5\right)}$$

Take the square root of -56.

$$t=\frac{-22±2\sqrt{14}i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$t=\frac{-22±2\sqrt{14}i}{-10}$$

Now solve the equation t=\frac{-22±2\sqrt{14}i}{-10} when ± is plus. Add -22 to 2i\sqrt{14}.

$$t=\frac{-22+2\sqrt{14}i}{-10}$$

Divide -22+2i\sqrt{14} by -10.

$$t=\frac{-\sqrt{14}i+11}{5}$$

Now solve the equation t=\frac{-22±2\sqrt{14}i}{-10} when ± is minus. Subtract 2i\sqrt{14} from -22.

$$t=\frac{-2\sqrt{14}i-22}{-10}$$

Divide -22-2i\sqrt{14} by -10.

$$t=\frac{11+\sqrt{14}i}{5}$$

The equation is now solved.

$$t=\frac{-\sqrt{14}i+11}{5} t=\frac{11+\sqrt{14}i}{5}$$

Steps for Completing the Square

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

$$22t-5t^{2}=27$$

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.

$$-5t^{2}+22t=27$$

Divide both sides by -5.

$$\frac{-5t^{2}+22t}{-5}=\frac{27}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$t^{2}+\frac{22}{-5}t=\frac{27}{-5}$$

Divide 22 by -5.

$$t^{2}-\frac{22}{5}t=\frac{27}{-5}$$

Divide 27 by -5.

$$t^{2}-\frac{22}{5}t=-\frac{27}{5}$$

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.

$$t^{2}-\frac{22}{5}t+\left(-\frac{11}{5}\right)^{2}=-\frac{27}{5}+\left(-\frac{11}{5}\right)^{2}$$

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

$$t^{2}-\frac{22}{5}t+\frac{121}{25}=-\frac{27}{5}+\frac{121}{25}$$

Add -\frac{27}{5} to \frac{121}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{22}{5}t+\frac{121}{25}=-\frac{14}{25}$$

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}.

$$\left(t-\frac{11}{5}\right)^{2}=-\frac{14}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{11}{5}\right)^{2}}=\sqrt{-\frac{14}{25}}$$

Simplify.

$$t-\frac{11}{5}=\frac{\sqrt{14}i}{5} t-\frac{11}{5}=-\frac{\sqrt{14}i}{5}$$

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

$$t=\frac{11+\sqrt{14}i}{5} t=\frac{-\sqrt{14}i+11}{5}$$