Solve -3x-18=-x^2

$$- 3 x - 18 = - x ^ { 2 }$$

Solve for x

$$x=-3$$

$$x=6$$

Steps Using Factoring

Add x^{2} to both sides.

$$-3x-18+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-3x-18=0$$

To solve the equation, factor x^{2}-3x-18 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.

$$a+b=-3 ab=-18$$

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 -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -3.

$$a=-6 b=3$$

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

$$\left(x-6\right)\left(x+3\right)$$

To find equation solutions, solve x-6=0 and x+3=0.

$$x=6 x=-3$$

Steps Using Factoring By Grouping

Add x^{2} to both sides.

$$-3x-18+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-3x-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-18\right)=-18$$

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 -18.

$$1,-18 2,-9 3,-6$$

Calculate the sum for each pair.

$$1-18=-17 2-9=-7 3-6=-3$$

The solution is the pair that gives sum -3.

$$a=-6 b=3$$

Rewrite x^{2}-3x-18 as \left(x^{2}-6x\right)+\left(3x-18\right).

$$\left(x^{2}-6x\right)+\left(3x-18\right)$$

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

$$x\left(x-6\right)+3\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+3\right)$$

To find equation solutions, solve x-6=0 and x+3=0.

$$x=6 x=-3$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

$$-3x-18+x^{2}=0$$

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.

$$x^{2}-3x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-18\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2}$$

Add 9 to 72.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-\left(-3\right)±9}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±9}{2}$$

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

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

Divide 12 by 2.

$$x=6$$

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

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

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=6 x=-3$$

Steps for Completing the Square

Add x^{2} to both sides.

$$-3x-18+x^{2}=0$$

Add 18 to both sides. Anything plus zero gives itself.

$$-3x+x^{2}=18$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}-3x=18$$

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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=18+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=18+\frac{9}{4}$$

Add 18 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{81}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{9}{2} x-\frac{3}{2}=-\frac{9}{2}$$

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

$$x=6 x=-3$$