Solve left(3-xright)^2=1-x^2/5

$${ \left(3-x \right) }^{ 2 } =1- \frac{ { x }^{ 2 } }{ 5 }$$

Solve for x (complex solution)

$$x=\frac{\sqrt{15}i}{6}+\frac{5}{2}\approx 2.5+0.645497224i$$

$$x=-\frac{\sqrt{15}i}{6}+\frac{5}{2}\approx 2.5-0.645497224i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5.

$$5\left(3-x\right)^{2}=5-x^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$5\left(9-6x+x^{2}\right)=5-x^{2}$$

Use the distributive property to multiply 5 by 9-6x+x^{2}.

$$45-30x+5x^{2}=5-x^{2}$$

Subtract 5 from both sides.

$$45-30x+5x^{2}-5=-x^{2}$$

Subtract 5 from 45 to get 40.

$$40-30x+5x^{2}=-x^{2}$$

Add x^{2} to both sides.

$$40-30x+5x^{2}+x^{2}=0$$

Combine 5x^{2} and x^{2} to get 6x^{2}.

$$40-30x+6x^{2}=0$$

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.

$$6x^{2}-30x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -30 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 6\times 40}}{2\times 6}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 6\times 40}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-30\right)±\sqrt{900-24\times 40}}{2\times 6}$$

Multiply -24 times 40.

$$x=\frac{-\left(-30\right)±\sqrt{900-960}}{2\times 6}$$

Add 900 to -960.

$$x=\frac{-\left(-30\right)±\sqrt{-60}}{2\times 6}$$

Take the square root of -60.

$$x=\frac{-\left(-30\right)±2\sqrt{15}i}{2\times 6}$$

The opposite of -30 is 30.

$$x=\frac{30±2\sqrt{15}i}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{30±2\sqrt{15}i}{12}$$

Now solve the equation x=\frac{30±2\sqrt{15}i}{12} when ± is plus. Add 30 to 2i\sqrt{15}.

$$x=\frac{30+2\sqrt{15}i}{12}$$

Divide 30+2i\sqrt{15} by 12.

$$x=\frac{\sqrt{15}i}{6}+\frac{5}{2}$$

Now solve the equation x=\frac{30±2\sqrt{15}i}{12} when ± is minus. Subtract 2i\sqrt{15} from 30.

$$x=\frac{-2\sqrt{15}i+30}{12}$$

Divide 30-2i\sqrt{15} by 12.

$$x=-\frac{\sqrt{15}i}{6}+\frac{5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{15}i}{6}+\frac{5}{2} x=-\frac{\sqrt{15}i}{6}+\frac{5}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 5.

$$5\left(3-x\right)^{2}=5-x^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$5\left(9-6x+x^{2}\right)=5-x^{2}$$

Use the distributive property to multiply 5 by 9-6x+x^{2}.

$$45-30x+5x^{2}=5-x^{2}$$

Add x^{2} to both sides.

$$45-30x+5x^{2}+x^{2}=5$$

Combine 5x^{2} and x^{2} to get 6x^{2}.

$$45-30x+6x^{2}=5$$

Subtract 45 from both sides.

$$-30x+6x^{2}=5-45$$

Subtract 45 from 5 to get -40.

$$-30x+6x^{2}=-40$$

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.

$$6x^{2}-30x=-40$$

Divide both sides by 6.

$$\frac{6x^{2}-30x}{6}=-\frac{40}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{30}{6}\right)x=-\frac{40}{6}$$

Divide -30 by 6.

$$x^{2}-5x=-\frac{40}{6}$$

Reduce the fraction \frac{-40}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-5x=-\frac{20}{3}$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-\frac{20}{3}+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-\frac{20}{3}+\frac{25}{4}$$

Add -\frac{20}{3} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=-\frac{5}{12}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{5}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{5}{12}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{15}i}{6} x-\frac{5}{2}=-\frac{\sqrt{15}i}{6}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{15}i}{6}+\frac{5}{2} x=-\frac{\sqrt{15}i}{6}+\frac{5}{2}$$