Solve (x-5)^2/2+x=3

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

Solve for x (complex solution)

$$x=4+\sqrt{3}i\approx 4+1.732050808i$$

$$x=-\sqrt{3}i+4\approx 4-1.732050808i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25+2x=6$$

Combine -10x and 2x to get -8x.

$$x^{2}-8x+25=6$$

Subtract 6 from both sides.

$$x^{2}-8x+25-6=0$$

Subtract 6 from 25 to get 19.

$$x^{2}-8x+19=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 19}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 19}}{2}$$

Multiply -4 times 19.

$$x=\frac{-\left(-8\right)±\sqrt{64-76}}{2}$$

Add 64 to -76.

$$x=\frac{-\left(-8\right)±\sqrt{-12}}{2}$$

Take the square root of -12.

$$x=\frac{-\left(-8\right)±2\sqrt{3}i}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{3}i}{2}$$

Now solve the equation x=\frac{8±2\sqrt{3}i}{2} when ± is plus. Add 8 to 2i\sqrt{3}.

$$x=\frac{8+2\sqrt{3}i}{2}$$

Divide 8+2i\sqrt{3} by 2.

$$x=4+\sqrt{3}i$$

Now solve the equation x=\frac{8±2\sqrt{3}i}{2} when ± is minus. Subtract 2i\sqrt{3} from 8.

$$x=\frac{-2\sqrt{3}i+8}{2}$$

Divide 8-2i\sqrt{3} by 2.

$$x=-\sqrt{3}i+4$$

The equation is now solved.

$$x=4+\sqrt{3}i x=-\sqrt{3}i+4$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25+2x=6$$

Combine -10x and 2x to get -8x.

$$x^{2}-8x+25=6$$

Subtract 25 from both sides.

$$x^{2}-8x=6-25$$

Subtract 25 from 6 to get -19.

$$x^{2}-8x=-19$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-19+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-19+16$$

Add -19 to 16.

$$x^{2}-8x+16=-3$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{-3}$$

Simplify.

$$x-4=\sqrt{3}i x-4=-\sqrt{3}i$$

Add 4 to both sides of the equation.

$$x=4+\sqrt{3}i x=-\sqrt{3}i+4$$