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

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

Solve for x (complex solution)

$$x=-\frac{\sqrt{26}i}{2}\approx -0-2.549509757i$$

$$x=\frac{\sqrt{26}i}{2}\approx 2.549509757i$$

Steps by Finding Square Root

Subtract 5 from 2 to get -3.

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

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

$$9-6x+x^{2}+\left(1-x-4\right)^{2}=5$$

Subtract 4 from 1 to get -3.

$$9-6x+x^{2}+\left(-3-x\right)^{2}=5$$

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

$$9-6x+x^{2}+9+6x+x^{2}=5$$

Add 9 and 9 to get 18.

$$18-6x+x^{2}+6x+x^{2}=5$$

Combine -6x and 6x to get 0.

$$18+x^{2}+x^{2}=5$$

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

$$18+2x^{2}=5$$

Subtract 18 from both sides.

$$2x^{2}=5-18$$

Subtract 18 from 5 to get -13.

$$2x^{2}=-13$$

Divide both sides by 2.

$$x^{2}=-\frac{13}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{26}i}{2} x=-\frac{\sqrt{26}i}{2}$$

Steps Using the Quadratic Formula

Subtract 5 from 2 to get -3.

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

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

$$9-6x+x^{2}+\left(1-x-4\right)^{2}=5$$

Subtract 4 from 1 to get -3.

$$9-6x+x^{2}+\left(-3-x\right)^{2}=5$$

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

$$9-6x+x^{2}+9+6x+x^{2}=5$$

Add 9 and 9 to get 18.

$$18-6x+x^{2}+6x+x^{2}=5$$

Combine -6x and 6x to get 0.

$$18+x^{2}+x^{2}=5$$

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

$$18+2x^{2}=5$$

Subtract 5 from both sides.

$$18+2x^{2}-5=0$$

Subtract 5 from 18 to get 13.

$$13+2x^{2}=0$$

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.

$$2x^{2}+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\times 13}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\times 13}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\times 13}}{2\times 2}$$

Multiply -8 times 13.

$$x=\frac{0±\sqrt{-104}}{2\times 2}$$

Take the square root of -104.

$$x=\frac{0±2\sqrt{26}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±2\sqrt{26}i}{4}$$

Now solve the equation x=\frac{0±2\sqrt{26}i}{4} when ± is plus.

$$x=\frac{\sqrt{26}i}{2}$$

Now solve the equation x=\frac{0±2\sqrt{26}i}{4} when ± is minus.

$$x=-\frac{\sqrt{26}i}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{26}i}{2} x=-\frac{\sqrt{26}i}{2}$$