Solve 10^2+x^2=10x+3

$${ 10 }^{ 2 } + { x }^{ 2 } =10x+3$$

Solve for x (complex solution)

$$x=5+6\sqrt{2}i\approx 5+8.485281374i$$

$$x=-6\sqrt{2}i+5\approx 5-8.485281374i$$

Steps Using the Quadratic Formula

Calculate 10 to the power of 2 and get 100.

$$100+x^{2}=10x+3$$

Subtract 10x from both sides.

$$100+x^{2}-10x=3$$

Subtract 3 from both sides.

$$100+x^{2}-10x-3=0$$

Subtract 3 from 100 to get 97.

$$97+x^{2}-10x=0$$

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.

$$x^{2}-10x+97=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 97 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 97}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 97}}{2}$$

Multiply -4 times 97.

$$x=\frac{-\left(-10\right)±\sqrt{100-388}}{2}$$

Add 100 to -388.

$$x=\frac{-\left(-10\right)±\sqrt{-288}}{2}$$

Take the square root of -288.

$$x=\frac{-\left(-10\right)±12\sqrt{2}i}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±12\sqrt{2}i}{2}$$

Now solve the equation x=\frac{10±12\sqrt{2}i}{2} when ± is plus. Add 10 to 12i\sqrt{2}.

$$x=\frac{10+12\sqrt{2}i}{2}$$

Divide 10+12i\sqrt{2} by 2.

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

Now solve the equation x=\frac{10±12\sqrt{2}i}{2} when ± is minus. Subtract 12i\sqrt{2} from 10.

$$x=\frac{-12\sqrt{2}i+10}{2}$$

Divide 10-12i\sqrt{2} by 2.

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

The equation is now solved.

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

Steps for Completing the Square

Calculate 10 to the power of 2 and get 100.

$$100+x^{2}=10x+3$$

Subtract 10x from both sides.

$$100+x^{2}-10x=3$$

Subtract 100 from both sides.

$$x^{2}-10x=3-100$$

Subtract 100 from 3 to get -97.

$$x^{2}-10x=-97$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-97+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-97+25$$

Add -97 to 25.

$$x^{2}-10x+25=-72$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=-72$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{-72}$$

Simplify.

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

Add 5 to both sides of the equation.

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