Solve 5=1-y^2

$$5 = 1 - y ^ { 2 }$$

Solve for y

$$y=-2i$$

$$y=2i$$

Steps by Finding Square Root

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

$$1-y^{2}=5$$

Subtract 1 from both sides.

$$-y^{2}=5-1$$

Subtract 1 from 5 to get 4.

$$-y^{2}=4$$

Divide both sides by -1.

$$y^{2}=-4$$

The equation is now solved.

$$y=2i y=-2i$$

Steps Using the Quadratic Formula

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

$$1-y^{2}=5$$

Subtract 5 from both sides.

$$1-y^{2}-5=0$$

Subtract 5 from 1 to get -4.

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

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.

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

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}.

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

Square 0.

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

Multiply -4 times -1.

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

Multiply 4 times -4.

$$y=\frac{0±\sqrt{-16}}{2\left(-1\right)}$$

Take the square root of -16.

$$y=\frac{0±4i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{0±4i}{-2}$$

Now solve the equation y=\frac{0±4i}{-2} when ± is plus.

$$y=-2i$$

Now solve the equation y=\frac{0±4i}{-2} when ± is minus.

$$y=2i$$

The equation is now solved.

$$y=-2i y=2i$$