Solve 0.5(1-x^2)=18

$$0.5(1- { x }^{ 2 } )=18$$

Solve for x (complex solution)

$$x=-i\sqrt{35}\approx -0-5.916079783i$$

$$x=i\sqrt{35}\approx 5.916079783i$$

Steps by Finding Square Root

Divide both sides by 0.5.

$$1-x^{2}=\frac{18}{0.5}$$

Expand \frac{18}{0.5} by multiplying both numerator and the denominator by 10.

$$1-x^{2}=\frac{180}{5}$$

Divide 180 by 5 to get 36.

$$1-x^{2}=36$$

Subtract 1 from both sides.

$$-x^{2}=36-1$$

Subtract 1 from 36 to get 35.

$$-x^{2}=35$$

Divide both sides by -1.

$$x^{2}=-35$$

The equation is now solved.

$$x=\sqrt{35}i x=-\sqrt{35}i$$

Steps Using the Quadratic Formula

Divide both sides by 0.5.

$$1-x^{2}=\frac{18}{0.5}$$

Expand \frac{18}{0.5} by multiplying both numerator and the denominator by 10.

$$1-x^{2}=\frac{180}{5}$$

Divide 180 by 5 to get 36.

$$1-x^{2}=36$$

Subtract 36 from both sides.

$$1-x^{2}-36=0$$

Subtract 36 from 1 to get -35.

$$-35-x^{2}=0$$

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.

$$-x^{2}-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\left(-35\right)}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\left(-35\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\left(-35\right)}}{2\left(-1\right)}$$

Multiply 4 times -35.

$$x=\frac{0±\sqrt{-140}}{2\left(-1\right)}$$

Take the square root of -140.

$$x=\frac{0±2\sqrt{35}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±2\sqrt{35}i}{-2}$$

Now solve the equation x=\frac{0±2\sqrt{35}i}{-2} when ± is plus.

$$x=-\sqrt{35}i$$

Now solve the equation x=\frac{0±2\sqrt{35}i}{-2} when ± is minus.

$$x=\sqrt{35}i$$

The equation is now solved.

$$x=-\sqrt{35}i x=\sqrt{35}i$$