Solve 27=x(x-3)/2

$$27= \frac{ x(x-3) }{ 2 }$$

Solve for x

$$x=-6$$

$$x=9$$

Steps Using the Quadratic Formula

Multiply both sides by 2.

$$27\times 2=x\left(x-3\right)$$

Multiply 27 and 2 to get 54.

$$54=x\left(x-3\right)$$

Use the distributive property to multiply x by x-3.

$$54=x^{2}-3x$$

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

$$x^{2}-3x=54$$

Subtract 54 from both sides.

$$x^{2}-3x-54=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -54 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-54\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-54\right)}}{2}$$

Multiply -4 times -54.

$$x=\frac{-\left(-3\right)±\sqrt{9+216}}{2}$$

Add 9 to 216.

$$x=\frac{-\left(-3\right)±\sqrt{225}}{2}$$

Take the square root of 225.

$$x=\frac{-\left(-3\right)±15}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±15}{2}$$

Now solve the equation x=\frac{3±15}{2} when ± is plus. Add 3 to 15.

$$x=\frac{18}{2}$$

Divide 18 by 2.

$$x=9$$

Now solve the equation x=\frac{3±15}{2} when ± is minus. Subtract 15 from 3.

$$x=-\frac{12}{2}$$

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=9 x=-6$$

Steps for Completing the Square

Multiply both sides by 2.

$$27\times 2=x\left(x-3\right)$$

Multiply 27 and 2 to get 54.

$$54=x\left(x-3\right)$$

Use the distributive property to multiply x by x-3.

$$54=x^{2}-3x$$

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

$$x^{2}-3x=54$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=54+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=54+\frac{9}{4}$$

Add 54 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{225}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{15}{2} x-\frac{3}{2}=-\frac{15}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=9 x=-6$$