Solve 0=-x^2-3x+54

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

Solve for x

$$x=-9$$

$$x=6$$

Steps Using Factoring By Grouping

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+54. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-54=-54$$

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 -54.

$$1,-54 2,-27 3,-18 6,-9$$

Calculate the sum for each pair.

$$1-54=-53 2-27=-25 3-18=-15 6-9=-3$$

The solution is the pair that gives sum -3.

$$a=6 b=-9$$

Rewrite -x^{2}-3x+54 as \left(-x^{2}+6x\right)+\left(-9x+54\right).

$$\left(-x^{2}+6x\right)+\left(-9x+54\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(-x+6\right)+9\left(-x+6\right)$$

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

$$\left(-x+6\right)\left(x+9\right)$$

To find equation solutions, solve -x+6=0 and x+9=0.

$$x=6 x=-9$$

Steps Using the Quadratic Formula

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

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

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}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 54}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 54}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 54}}{2\left(-1\right)}$$

Multiply 4 times 54.

$$x=\frac{-\left(-3\right)±\sqrt{9+216}}{2\left(-1\right)}$$

Add 9 to 216.

$$x=\frac{-\left(-3\right)±\sqrt{225}}{2\left(-1\right)}$$

Take the square root of 225.

$$x=\frac{-\left(-3\right)±15}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±15}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±15}{-2}$$

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

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

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

Subtract 54 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

$$\frac{-x^{2}-3x}{-1}=-\frac{54}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{3}{-1}\right)x=-\frac{54}{-1}$$

Divide -3 by -1.

$$x^{2}+3x=-\frac{54}{-1}$$

Divide -54 by -1.

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

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.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=6 x=-9$$