Solve -3/4t^2-9/4t+3-3/4t-3=3

$$- \frac { 3 } { 4 } t ^ { 2 } - \frac { 9 } { 4 } t + 3 - \frac { 3 } { 4 } t - 3 = 3$$

Solve for t

$$t=-2$$

Steps Using the Quadratic Formula

Combine -\frac{9}{4}t and -\frac{3}{4}t to get -3t.

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

Subtract 3 from 3 to get 0.

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

Subtract 3 from both sides.

$$-\frac{3}{4}t^{2}-3t-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{4} for a, -3 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-\frac{3}{4}\right)\left(-3\right)}}{2\left(-\frac{3}{4}\right)}$$

Square -3.

$$t=\frac{-\left(-3\right)±\sqrt{9-4\left(-\frac{3}{4}\right)\left(-3\right)}}{2\left(-\frac{3}{4}\right)}$$

Multiply -4 times -\frac{3}{4}.

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

Multiply 3 times -3.

$$t=\frac{-\left(-3\right)±\sqrt{9-9}}{2\left(-\frac{3}{4}\right)}$$

Add 9 to -9.

$$t=\frac{-\left(-3\right)±\sqrt{0}}{2\left(-\frac{3}{4}\right)}$$

Take the square root of 0.

$$t=-\frac{-3}{2\left(-\frac{3}{4}\right)}$$

The opposite of -3 is 3.

$$t=\frac{3}{2\left(-\frac{3}{4}\right)}$$

Multiply 2 times -\frac{3}{4}.

$$t=\frac{3}{-\frac{3}{2}}$$

Divide 3 by -\frac{3}{2} by multiplying 3 by the reciprocal of -\frac{3}{2}.

$$t=-2$$

Steps for Completing the Square

Combine -\frac{9}{4}t and -\frac{3}{4}t to get -3t.

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

Subtract 3 from 3 to get 0.

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

Divide both sides of the equation by -\frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

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

Dividing by -\frac{3}{4} undoes the multiplication by -\frac{3}{4}.

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

Divide -3 by -\frac{3}{4} by multiplying -3 by the reciprocal of -\frac{3}{4}.

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

Divide 3 by -\frac{3}{4} by multiplying 3 by the reciprocal of -\frac{3}{4}.

$$t^{2}+4t=-4$$

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

$$t^{2}+4t+2^{2}=-4+2^{2}$$

Square 2.

$$t^{2}+4t+4=-4+4$$

Add -4 to 4.

$$t^{2}+4t+4=0$$

Factor t^{2}+4t+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}.

$$\left(t+2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+2\right)^{2}}=\sqrt{0}$$

Simplify.

$$t+2=0 t+2=0$$

Subtract 2 from both sides of the equation.

$$t=-2 t=-2$$

The equation is now solved. Solutions are the same.

$$t=-2$$