Solve -t^2-4t+3=t/2

$$- t ^ { 2 } - 4 t + 3 = \frac { t } { 2 }$$

Solve for t

$$t=\frac{\sqrt{129}-9}{4}\approx 0.589454173$$

$$t=\frac{-\sqrt{129}-9}{4}\approx -5.089454173$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$-2t^{2}-8t+6=t$$

Subtract t from both sides.

$$-2t^{2}-8t+6-t=0$$

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

$$-2t^{2}-9t+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -9 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

Square -9.

$$t=\frac{-\left(-9\right)±\sqrt{81-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$t=\frac{-\left(-9\right)±\sqrt{81+8\times 6}}{2\left(-2\right)}$$

Multiply 8 times 6.

$$t=\frac{-\left(-9\right)±\sqrt{81+48}}{2\left(-2\right)}$$

Add 81 to 48.

$$t=\frac{-\left(-9\right)±\sqrt{129}}{2\left(-2\right)}$$

The opposite of -9 is 9.

$$t=\frac{9±\sqrt{129}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$t=\frac{9±\sqrt{129}}{-4}$$

Now solve the equation t=\frac{9±\sqrt{129}}{-4} when ± is plus. Add 9 to \sqrt{129}.

$$t=\frac{\sqrt{129}+9}{-4}$$

Divide 9+\sqrt{129} by -4.

$$t=\frac{-\sqrt{129}-9}{4}$$

Now solve the equation t=\frac{9±\sqrt{129}}{-4} when ± is minus. Subtract \sqrt{129} from 9.

$$t=\frac{9-\sqrt{129}}{-4}$$

Divide 9-\sqrt{129} by -4.

$$t=\frac{\sqrt{129}-9}{4}$$

The equation is now solved.

$$t=\frac{-\sqrt{129}-9}{4} t=\frac{\sqrt{129}-9}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$-2t^{2}-8t+6=t$$

Subtract t from both sides.

$$-2t^{2}-8t+6-t=0$$

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

$$-2t^{2}-9t+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-2t^{2}-9t=-6$$

Divide both sides by -2.

$$\frac{-2t^{2}-9t}{-2}=-\frac{6}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$t^{2}+\left(-\frac{9}{-2}\right)t=-\frac{6}{-2}$$

Divide -9 by -2.

$$t^{2}+\frac{9}{2}t=-\frac{6}{-2}$$

Divide -6 by -2.

$$t^{2}+\frac{9}{2}t=3$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{9}{2}t+\left(\frac{9}{4}\right)^{2}=3+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{9}{2}t+\frac{81}{16}=3+\frac{81}{16}$$

Add 3 to \frac{81}{16}.

$$t^{2}+\frac{9}{2}t+\frac{81}{16}=\frac{129}{16}$$

Factor t^{2}+\frac{9}{2}t+\frac{81}{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}.

$$\left(t+\frac{9}{4}\right)^{2}=\frac{129}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{9}{4}\right)^{2}}=\sqrt{\frac{129}{16}}$$

Simplify.

$$t+\frac{9}{4}=\frac{\sqrt{129}}{4} t+\frac{9}{4}=-\frac{\sqrt{129}}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$t=\frac{\sqrt{129}-9}{4} t=\frac{-\sqrt{129}-9}{4}$$