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

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

Solve for t

$$t=1$$

$$t=-1$$

Steps by Finding Square Root

Combine t and -t to get 0.

$$1-t^{2}=1\times 0$$

Multiply 1 and 0 to get 0.

$$1-t^{2}=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-t^{2}=-1$$

Divide both sides by -1.

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

Divide -1 by -1 to get 1.

$$t^{2}=1$$

Take the square root of both sides of the equation.

$$t=1 t=-1$$

Steps Using the Quadratic Formula

Combine t and -t to get 0.

$$1-t^{2}=1\times 0$$

Multiply 1 and 0 to get 0.

$$1-t^{2}=0$$

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.

$$-t^{2}+1=0$$

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

$$t=\frac{0±\sqrt{0^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Square 0.

$$t=\frac{0±\sqrt{-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$t=\frac{0±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$t=\frac{0±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$t=\frac{0±2}{-2}$$

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

$$t=-1$$

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

$$t=1$$

The equation is now solved.

$$t=-1 t=1$$