Solve 2t^2+9+t=0quadt=

$$2 t ^ { 2 } + 9 + t = 0 \quad t =$$

Solve for t

$$t=\frac{-1+\sqrt{71}i}{4}\approx -0.25+2.106537443i$$

$$t=\frac{-\sqrt{71}i-1}{4}\approx -0.25-2.106537443i$$

Steps Using the Quadratic Formula

Anything times zero gives zero.

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

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.

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

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

$$t=\frac{-1±\sqrt{1^{2}-4\times 2\times 9}}{2\times 2}$$

Square 1.

$$t=\frac{-1±\sqrt{1-4\times 2\times 9}}{2\times 2}$$

Multiply -4 times 2.

$$t=\frac{-1±\sqrt{1-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$t=\frac{-1±\sqrt{1-72}}{2\times 2}$$

Add 1 to -72.

$$t=\frac{-1±\sqrt{-71}}{2\times 2}$$

Take the square root of -71.

$$t=\frac{-1±\sqrt{71}i}{2\times 2}$$

Multiply 2 times 2.

$$t=\frac{-1±\sqrt{71}i}{4}$$

Now solve the equation t=\frac{-1±\sqrt{71}i}{4} when ± is plus. Add -1 to i\sqrt{71}.

$$t=\frac{-1+\sqrt{71}i}{4}$$

Now solve the equation t=\frac{-1±\sqrt{71}i}{4} when ± is minus. Subtract i\sqrt{71} from -1.

$$t=\frac{-\sqrt{71}i-1}{4}$$

The equation is now solved.

$$t=\frac{-1+\sqrt{71}i}{4} t=\frac{-\sqrt{71}i-1}{4}$$

Steps for Completing the Square

Anything times zero gives zero.

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

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

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{1}{2}t+\left(\frac{1}{4}\right)^{2}=-\frac{9}{2}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{1}{2}t+\frac{1}{16}=-\frac{9}{2}+\frac{1}{16}$$

Add -\frac{9}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{1}{2}t+\frac{1}{16}=-\frac{71}{16}$$

Factor t^{2}+\frac{1}{2}t+\frac{1}{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}.

$$\left(t+\frac{1}{4}\right)^{2}=-\frac{71}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{71}{16}}$$

Simplify.

$$t+\frac{1}{4}=\frac{\sqrt{71}i}{4} t+\frac{1}{4}=-\frac{\sqrt{71}i}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$t=\frac{-1+\sqrt{71}i}{4} t=\frac{-\sqrt{71}i-1}{4}$$