Solve x^2-3x=27+3x

$$x ^ { 2 } - 3 x = 27 + 3 x$$

Solve for x

$$x=-3$$

$$x=9$$

Steps Using Factoring

Subtract 27 from both sides.

$$x^{2}-3x-27=3x$$

Subtract 3x from both sides.

$$x^{2}-3x-27-3x=0$$

Combine -3x and -3x to get -6x.

$$x^{2}-6x-27=0$$

To solve the equation, factor x^{2}-6x-27 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-6 ab=-27$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -27.

$$1,-27 3,-9$$

Calculate the sum for each pair.

$$1-27=-26 3-9=-6$$

The solution is the pair that gives sum -6.

$$a=-9 b=3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-9\right)\left(x+3\right)$$

To find equation solutions, solve x-9=0 and x+3=0.

$$x=9 x=-3$$

Steps Using Factoring By Grouping

Subtract 27 from both sides.

$$x^{2}-3x-27=3x$$

Subtract 3x from both sides.

$$x^{2}-3x-27-3x=0$$

Combine -3x and -3x to get -6x.

$$x^{2}-6x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\left(-27\right)=-27$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -27.

$$1,-27 3,-9$$

Calculate the sum for each pair.

$$1-27=-26 3-9=-6$$

The solution is the pair that gives sum -6.

$$a=-9 b=3$$

Rewrite x^{2}-6x-27 as \left(x^{2}-9x\right)+\left(3x-27\right).

$$\left(x^{2}-9x\right)+\left(3x-27\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(x-9\right)+3\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x+3\right)$$

To find equation solutions, solve x-9=0 and x+3=0.

$$x=9 x=-3$$

Steps Using the Quadratic Formula

Subtract 27 from both sides.

$$x^{2}-3x-27=3x$$

Subtract 3x from both sides.

$$x^{2}-3x-27-3x=0$$

Combine -3x and -3x to get -6x.

$$x^{2}-6x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-27\right)}}{2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-27\right)}}{2}$$

Multiply -4 times -27.

$$x=\frac{-\left(-6\right)±\sqrt{36+108}}{2}$$

Add 36 to 108.

$$x=\frac{-\left(-6\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$x=\frac{-\left(-6\right)±12}{2}$$

The opposite of -6 is 6.

$$x=\frac{6±12}{2}$$

Now solve the equation x=\frac{6±12}{2} when ± is plus. Add 6 to 12.

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

Divide 18 by 2.

$$x=9$$

Now solve the equation x=\frac{6±12}{2} when ± is minus. Subtract 12 from 6.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=9 x=-3$$

Steps for Completing the Square

Subtract 3x from both sides.

$$x^{2}-3x-3x=27$$

Combine -3x and -3x to get -6x.

$$x^{2}-6x=27$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=27+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=27+9$$

Add 27 to 9.

$$x^{2}-6x+9=36$$

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}.

$$\left(x-3\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{36}$$

Simplify.

$$x-3=6 x-3=-6$$

Add 3 to both sides of the equation.

$$x=9 x=-3$$