Solve t^2-4=0

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

Solve for t

$$t=2$$

$$t=-2$$

Steps Using Difference of Squares

Consider t^{2}-4. Rewrite t^{2}-4 as t^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(t-2\right)\left(t+2\right)=0$$

To find equation solutions, solve t-2=0 and t+2=0.

$$t=2 t=-2$$

Steps by Finding Square Root

Add 4 to both sides. Anything plus zero gives itself.

$$t^{2}=4$$

Take the square root of both sides of the equation.

$$t=2 t=-2$$

Steps Using the Quadratic Formula

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

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

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

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

Square 0.

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

Multiply -4 times -4.

$$t=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$t=\frac{0±4}{2}$$

Now solve the equation t=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$t=2$$

Now solve the equation t=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$t=-2$$

The equation is now solved.

$$t=2 t=-2$$