Solve left(-1-xright)^2+left(4-(-x+3)right)^2=5/3

$${ \left(-1-x \right) }^{ 2 } + { \left(4-(-x+3) \right) }^{ 2 } = \frac{ 5 }{ 3 }$$

Solve for x

$$x=-\frac{\sqrt{30}}{6}-1\approx -1.912870929$$

$$x=\frac{\sqrt{30}}{6}-1\approx -0.087129071$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-1-x\right)^{2}.

$$1+2x+x^{2}+\left(4-\left(-x+3\right)\right)^{2}=\frac{5}{3}$$

To find the opposite of -x+3, find the opposite of each term.

$$1+2x+x^{2}+\left(4-\left(-x\right)-3\right)^{2}=\frac{5}{3}$$

Subtract 3 from 4 to get 1.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1-\left(-x\right)\right)^{2}.

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

Calculate -\left(-x\right) to the power of 2 and get \left(-x\right)^{2}.

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

Add 1 and 1 to get 2.

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

Calculate -x to the power of 2 and get x^{2}.

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

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

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

Subtract \frac{5}{3} from both sides.

$$2+2x+2x^{2}+2\left(-\left(-x\right)\right)-\frac{5}{3}=0$$

Subtract \frac{5}{3} from 2 to get \frac{1}{3}.

$$\frac{1}{3}+2x+2x^{2}+2\left(-\left(-x\right)\right)=0$$

Multiply 2 and -1 to get -2.

$$\frac{1}{3}+2x+2x^{2}-2\left(-1\right)x=0$$

Multiply -2 and -1 to get 2.

$$\frac{1}{3}+2x+2x^{2}+2x=0$$

Combine 2x and 2x to get 4x.

$$\frac{1}{3}+4x+2x^{2}=0$$

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.

$$2x^{2}+4x+\frac{1}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 4 for b, and \frac{1}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\times \frac{1}{3}}}{2\times 2}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 2\times \frac{1}{3}}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-4±\sqrt{16-8\times \frac{1}{3}}}{2\times 2}$$

Multiply -8 times \frac{1}{3}.

$$x=\frac{-4±\sqrt{16-\frac{8}{3}}}{2\times 2}$$

Add 16 to -\frac{8}{3}.

$$x=\frac{-4±\sqrt{\frac{40}{3}}}{2\times 2}$$

Take the square root of \frac{40}{3}.

$$x=\frac{-4±\frac{2\sqrt{30}}{3}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±\frac{2\sqrt{30}}{3}}{4}$$

Now solve the equation x=\frac{-4±\frac{2\sqrt{30}}{3}}{4} when ± is plus. Add -4 to \frac{2\sqrt{30}}{3}.

$$x=\frac{\frac{2\sqrt{30}}{3}-4}{4}$$

Divide -4+\frac{2\sqrt{30}}{3} by 4.

$$x=\frac{\sqrt{30}}{6}-1$$

Now solve the equation x=\frac{-4±\frac{2\sqrt{30}}{3}}{4} when ± is minus. Subtract \frac{2\sqrt{30}}{3} from -4.

$$x=\frac{-\frac{2\sqrt{30}}{3}-4}{4}$$

Divide -4-\frac{2\sqrt{30}}{3} by 4.

$$x=-\frac{\sqrt{30}}{6}-1$$

The equation is now solved.

$$x=\frac{\sqrt{30}}{6}-1 x=-\frac{\sqrt{30}}{6}-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-1-x\right)^{2}.

$$1+2x+x^{2}+\left(4-\left(-x+3\right)\right)^{2}=\frac{5}{3}$$

To find the opposite of -x+3, find the opposite of each term.

$$1+2x+x^{2}+\left(4-\left(-x\right)-3\right)^{2}=\frac{5}{3}$$

Subtract 3 from 4 to get 1.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1-\left(-x\right)\right)^{2}.

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

Calculate -\left(-x\right) to the power of 2 and get \left(-x\right)^{2}.

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

Add 1 and 1 to get 2.

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

Calculate -x to the power of 2 and get x^{2}.

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

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

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

Subtract 2 from both sides.

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

Subtract 2 from \frac{5}{3} to get -\frac{1}{3}.

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

Multiply 2 and -1 to get -2.

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

Multiply -2 and -1 to get 2.

$$2x+2x^{2}+2x=-\frac{1}{3}$$

Combine 2x and 2x to get 4x.

$$4x+2x^{2}=-\frac{1}{3}$$

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.

$$2x^{2}+4x=-\frac{1}{3}$$

Divide both sides by 2.

$$\frac{2x^{2}+4x}{2}=-\frac{\frac{1}{3}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{4}{2}x=-\frac{\frac{1}{3}}{2}$$

Divide 4 by 2.

$$x^{2}+2x=-\frac{\frac{1}{3}}{2}$$

Divide -\frac{1}{3} by 2.

$$x^{2}+2x=-\frac{1}{6}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=-\frac{1}{6}+1^{2}$$

Square 1.

$$x^{2}+2x+1=-\frac{1}{6}+1$$

Add -\frac{1}{6} to 1.

$$x^{2}+2x+1=\frac{5}{6}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{5}{6}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{5}{6}}$$

Simplify.

$$x+1=\frac{\sqrt{30}}{6} x+1=-\frac{\sqrt{30}}{6}$$

Subtract 1 from both sides of the equation.

$$x=\frac{\sqrt{30}}{6}-1 x=-\frac{\sqrt{30}}{6}-1$$