Solve .9x(x-6)=81

$$.9 x ( x - 6 ) = 81$$

Solve for x

$$x=3\sqrt{11}+3\approx 12.949874371$$

$$x=3-3\sqrt{11}\approx -6.949874371$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 0.9x by x-6.

$$0.9x^{2}-5.4x=81$$

Subtract 81 from both sides.

$$0.9x^{2}-5.4x-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 0.9 for a, -5.4 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5.4\right)±\sqrt{\left(-5.4\right)^{2}-4\times 0.9\left(-81\right)}}{2\times 0.9}$$

Square -5.4 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-5.4\right)±\sqrt{29.16-4\times 0.9\left(-81\right)}}{2\times 0.9}$$

Multiply -4 times 0.9.

$$x=\frac{-\left(-5.4\right)±\sqrt{29.16-3.6\left(-81\right)}}{2\times 0.9}$$

Multiply -3.6 times -81.

$$x=\frac{-\left(-5.4\right)±\sqrt{29.16+291.6}}{2\times 0.9}$$

Add 29.16 to 291.6 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-5.4\right)±\sqrt{320.76}}{2\times 0.9}$$

Take the square root of 320.76.

$$x=\frac{-\left(-5.4\right)±\frac{27\sqrt{11}}{5}}{2\times 0.9}$$

The opposite of -5.4 is 5.4.

$$x=\frac{5.4±\frac{27\sqrt{11}}{5}}{2\times 0.9}$$

Multiply 2 times 0.9.

$$x=\frac{5.4±\frac{27\sqrt{11}}{5}}{1.8}$$

Now solve the equation x=\frac{5.4±\frac{27\sqrt{11}}{5}}{1.8} when ± is plus. Add 5.4 to \frac{27\sqrt{11}}{5}.

$$x=\frac{27\sqrt{11}+27}{1.8\times 5}$$

Divide \frac{27+27\sqrt{11}}{5} by 1.8 by multiplying \frac{27+27\sqrt{11}}{5} by the reciprocal of 1.8.

$$x=3\sqrt{11}+3$$

Now solve the equation x=\frac{5.4±\frac{27\sqrt{11}}{5}}{1.8} when ± is minus. Subtract \frac{27\sqrt{11}}{5} from 5.4.

$$x=\frac{27-27\sqrt{11}}{1.8\times 5}$$

Divide \frac{27-27\sqrt{11}}{5} by 1.8 by multiplying \frac{27-27\sqrt{11}}{5} by the reciprocal of 1.8.

$$x=3-3\sqrt{11}$$

The equation is now solved.

$$x=3\sqrt{11}+3 x=3-3\sqrt{11}$$

Steps for Completing the Square

Use the distributive property to multiply 0.9x by x-6.

$$0.9x^{2}-5.4x=81$$

Divide both sides of the equation by 0.9, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{0.9x^{2}-5.4x}{0.9}=\frac{81}{0.9}$$

Dividing by 0.9 undoes the multiplication by 0.9.

$$x^{2}+\left(-\frac{5.4}{0.9}\right)x=\frac{81}{0.9}$$

Divide -5.4 by 0.9 by multiplying -5.4 by the reciprocal of 0.9.

$$x^{2}-6x=\frac{81}{0.9}$$

Divide 81 by 0.9 by multiplying 81 by the reciprocal of 0.9.

$$x^{2}-6x=90$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=90+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=90+9$$

Add 90 to 9.

$$x^{2}-6x+9=99$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=99$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{99}$$

Simplify.

$$x-3=3\sqrt{11} x-3=-3\sqrt{11}$$

Add 3 to both sides of the equation.

$$x=3\sqrt{11}+3 x=3-3\sqrt{11}$$