Solve 27=6/100(x-10)^2

$$27 = \frac { 6 } { 100 } ( x - 10 ) ^ { 2 }$$

Solve for x

$$x=15\sqrt{2}+10\approx 31.213203436$$

$$x=10-15\sqrt{2}\approx -11.213203436$$

Steps Using the Quadratic Formula

Reduce the fraction \frac{6}{100} to lowest terms by extracting and canceling out 2.

$$27=\frac{3}{50}\left(x-10\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-10\right)^{2}.

$$27=\frac{3}{50}\left(x^{2}-20x+100\right)$$

Use the distributive property to multiply \frac{3}{50} by x^{2}-20x+100.

$$27=\frac{3}{50}x^{2}-\frac{6}{5}x+6$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{3}{50}x^{2}-\frac{6}{5}x+6=27$$

Subtract 27 from both sides.

$$\frac{3}{50}x^{2}-\frac{6}{5}x+6-27=0$$

Subtract 27 from 6 to get -21.

$$\frac{3}{50}x^{2}-\frac{6}{5}x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{50} for a, -\frac{6}{5} for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\left(-\frac{6}{5}\right)^{2}-4\times \frac{3}{50}\left(-21\right)}}{2\times \frac{3}{50}}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{36}{25}-4\times \frac{3}{50}\left(-21\right)}}{2\times \frac{3}{50}}$$

Multiply -4 times \frac{3}{50}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{36}{25}-\frac{6}{25}\left(-21\right)}}{2\times \frac{3}{50}}$$

Multiply -\frac{6}{25} times -21.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{36+126}{25}}}{2\times \frac{3}{50}}$$

Add \frac{36}{25} to \frac{126}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{162}{25}}}{2\times \frac{3}{50}}$$

Take the square root of \frac{162}{25}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\frac{9\sqrt{2}}{5}}{2\times \frac{3}{50}}$$

The opposite of -\frac{6}{5} is \frac{6}{5}.

$$x=\frac{\frac{6}{5}±\frac{9\sqrt{2}}{5}}{2\times \frac{3}{50}}$$

Multiply 2 times \frac{3}{50}.

$$x=\frac{\frac{6}{5}±\frac{9\sqrt{2}}{5}}{\frac{3}{25}}$$

Now solve the equation x=\frac{\frac{6}{5}±\frac{9\sqrt{2}}{5}}{\frac{3}{25}} when ± is plus. Add \frac{6}{5} to \frac{9\sqrt{2}}{5}.

$$x=\frac{9\sqrt{2}+6}{\frac{3}{25}\times 5}$$

Divide \frac{6+9\sqrt{2}}{5} by \frac{3}{25} by multiplying \frac{6+9\sqrt{2}}{5} by the reciprocal of \frac{3}{25}.

$$x=15\sqrt{2}+10$$

Now solve the equation x=\frac{\frac{6}{5}±\frac{9\sqrt{2}}{5}}{\frac{3}{25}} when ± is minus. Subtract \frac{9\sqrt{2}}{5} from \frac{6}{5}.

$$x=\frac{6-9\sqrt{2}}{\frac{3}{25}\times 5}$$

Divide \frac{6-9\sqrt{2}}{5} by \frac{3}{25} by multiplying \frac{6-9\sqrt{2}}{5} by the reciprocal of \frac{3}{25}.

$$x=10-15\sqrt{2}$$

The equation is now solved.

$$x=15\sqrt{2}+10 x=10-15\sqrt{2}$$

Steps for Completing the Square

Reduce the fraction \frac{6}{100} to lowest terms by extracting and canceling out 2.

$$27=\frac{3}{50}\left(x-10\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-10\right)^{2}.

$$27=\frac{3}{50}\left(x^{2}-20x+100\right)$$

Use the distributive property to multiply \frac{3}{50} by x^{2}-20x+100.

$$27=\frac{3}{50}x^{2}-\frac{6}{5}x+6$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{3}{50}x^{2}-\frac{6}{5}x+6=27$$

Subtract 6 from both sides.

$$\frac{3}{50}x^{2}-\frac{6}{5}x=27-6$$

Subtract 6 from 27 to get 21.

$$\frac{3}{50}x^{2}-\frac{6}{5}x=21$$

Divide both sides of the equation by \frac{3}{50}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{50}x^{2}-\frac{6}{5}x}{\frac{3}{50}}=\frac{21}{\frac{3}{50}}$$

Dividing by \frac{3}{50} undoes the multiplication by \frac{3}{50}.

$$x^{2}+\left(-\frac{\frac{6}{5}}{\frac{3}{50}}\right)x=\frac{21}{\frac{3}{50}}$$

Divide -\frac{6}{5} by \frac{3}{50} by multiplying -\frac{6}{5} by the reciprocal of \frac{3}{50}.

$$x^{2}-20x=\frac{21}{\frac{3}{50}}$$

Divide 21 by \frac{3}{50} by multiplying 21 by the reciprocal of \frac{3}{50}.

$$x^{2}-20x=350$$

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.

$$x^{2}-20x+\left(-10\right)^{2}=350+\left(-10\right)^{2}$$

Square -10.

$$x^{2}-20x+100=350+100$$

Add 350 to 100.

$$x^{2}-20x+100=450$$

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}.

$$\left(x-10\right)^{2}=450$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-10\right)^{2}}=\sqrt{450}$$

Simplify.

$$x-10=15\sqrt{2} x-10=-15\sqrt{2}$$

Add 10 to both sides of the equation.

$$x=15\sqrt{2}+10 x=10-15\sqrt{2}$$