Solve 3=-2^2-2t(-2)+t^2-1

$$3 = { -2 }^{ 2 } -2t(-2)+ { t }^{ 2 } -1$$

Solve for t

$$t=-4$$

$$t=0$$

Steps Using Factoring

Calculate -2 to the power of 2 and get 4.

$$3=4-2t\left(-2\right)+t^{2}-1$$

Multiply 2 and -2 to get -4.

$$3=4-\left(-4t\right)+t^{2}-1$$

The opposite of -4t is 4t.

$$3=4+4t+t^{2}-1$$

Subtract 1 from 4 to get 3.

$$3=3+4t+t^{2}$$

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

$$3+4t+t^{2}=3$$

Subtract 3 from both sides.

$$3+4t+t^{2}-3=0$$

Subtract 3 from 3 to get 0.

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

Factor out t.

$$t\left(4+t\right)=0$$

To find equation solutions, solve t=0 and 4+t=0.

$$t=0 t=-4$$

Steps Using the Quadratic Formula

Calculate -2 to the power of 2 and get 4.

$$3=4-2t\left(-2\right)+t^{2}-1$$

Multiply 2 and -2 to get -4.

$$3=4-\left(-4t\right)+t^{2}-1$$

The opposite of -4t is 4t.

$$3=4+4t+t^{2}-1$$

Subtract 1 from 4 to get 3.

$$3=3+4t+t^{2}$$

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

$$3+4t+t^{2}=3$$

Subtract 3 from both sides.

$$3+4t+t^{2}-3=0$$

Subtract 3 from 3 to get 0.

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

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.

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

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

$$t=\frac{-4±\sqrt{4^{2}}}{2}$$

Take the square root of 4^{2}.

$$t=\frac{-4±4}{2}$$

Now solve the equation t=\frac{-4±4}{2} when ± is plus. Add -4 to 4.

$$t=\frac{0}{2}$$

Divide 0 by 2.

$$t=0$$

Now solve the equation t=\frac{-4±4}{2} when ± is minus. Subtract 4 from -4.

$$t=-\frac{8}{2}$$

Divide -8 by 2.

$$t=-4$$

The equation is now solved.

$$t=0 t=-4$$

Steps for Completing the Square

Calculate -2 to the power of 2 and get 4.

$$3=4-2t\left(-2\right)+t^{2}-1$$

Multiply 2 and -2 to get -4.

$$3=4-\left(-4t\right)+t^{2}-1$$

The opposite of -4t is 4t.

$$3=4+4t+t^{2}-1$$

Subtract 1 from 4 to get 3.

$$3=3+4t+t^{2}$$

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

$$3+4t+t^{2}=3$$

Subtract 3 from both sides.

$$3+4t+t^{2}-3=0$$

Subtract 3 from 3 to get 0.

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

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.

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

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.

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

Square 2.

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

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(t+2\right)^{2}}=\sqrt{4}$$

Simplify.

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

Subtract 2 from both sides of the equation.

$$t=0 t=-4$$