Solve x^2+97=20x

$${ x }^{ 2 } +97 = 20x$$

Solve for x

$$x=\sqrt{3}+10\approx 11.732050808$$

$$x=10-\sqrt{3}\approx 8.267949192$$

Steps Using the Quadratic Formula

Subtract 20x from both sides.

$$x^{2}+97-20x=0$$

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.

$$x^{2}-20x+97=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -20 for b, and 97 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 97}}{2}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 97}}{2}$$

Multiply -4 times 97.

$$x=\frac{-\left(-20\right)±\sqrt{400-388}}{2}$$

Add 400 to -388.

$$x=\frac{-\left(-20\right)±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{-\left(-20\right)±2\sqrt{3}}{2}$$

The opposite of -20 is 20.

$$x=\frac{20±2\sqrt{3}}{2}$$

Now solve the equation x=\frac{20±2\sqrt{3}}{2} when ± is plus. Add 20 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+20}{2}$$

Divide 20+2\sqrt{3} by 2.

$$x=\sqrt{3}+10$$

Now solve the equation x=\frac{20±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 20.

$$x=\frac{20-2\sqrt{3}}{2}$$

Divide 20-2\sqrt{3} by 2.

$$x=10-\sqrt{3}$$

The equation is now solved.

$$x=\sqrt{3}+10 x=10-\sqrt{3}$$

Steps for Completing the Square

Subtract 20x from both sides.

$$x^{2}+97-20x=0$$

Subtract 97 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-20x=-97$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-20x+\left(-10\right)^{2}=-97+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=-97+100$$

Add -97 to 100.

$$x^{2}-20x+100=3$$

Factor x^{2}-20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-10\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-10\right)^{2}}=\sqrt{3}$$

Simplify.

$$x-10=\sqrt{3} x-10=-\sqrt{3}$$

Add 10 to both sides of the equation.

$$x=\sqrt{3}+10 x=10-\sqrt{3}$$