Solve 20=-12q-q^2

$$20 = - 12 q - q ^ { 2 }$$

Solve for q

$$q=-10$$

$$q=-2$$

Steps Using Factoring By Grouping

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

$$-12q-q^{2}=20$$

Subtract 20 from both sides.

$$-12q-q^{2}-20=0$$

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

$$-q^{2}-12q-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -q^{2}+aq+bq-20. To find a and b, set up a system to be solved.

$$a+b=-12 ab=-\left(-20\right)=20$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -12.

$$a=-2 b=-10$$

Rewrite -q^{2}-12q-20 as \left(-q^{2}-2q\right)+\left(-10q-20\right).

$$\left(-q^{2}-2q\right)+\left(-10q-20\right)$$

Factor out q in the first and 10 in the second group.

$$q\left(-q-2\right)+10\left(-q-2\right)$$

Factor out common term -q-2 by using distributive property.

$$\left(-q-2\right)\left(q+10\right)$$

To find equation solutions, solve -q-2=0 and q+10=0.

$$q=-2 q=-10$$

Steps Using the Quadratic Formula

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

$$-12q-q^{2}=20$$

Subtract 20 from both sides.

$$-12q-q^{2}-20=0$$

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.

$$-q^{2}-12q-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -12 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Square -12.

$$q=\frac{-\left(-12\right)±\sqrt{144-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$q=\frac{-\left(-12\right)±\sqrt{144+4\left(-20\right)}}{2\left(-1\right)}$$

Multiply 4 times -20.

$$q=\frac{-\left(-12\right)±\sqrt{144-80}}{2\left(-1\right)}$$

Add 144 to -80.

$$q=\frac{-\left(-12\right)±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$q=\frac{-\left(-12\right)±8}{2\left(-1\right)}$$

The opposite of -12 is 12.

$$q=\frac{12±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$q=\frac{12±8}{-2}$$

Now solve the equation q=\frac{12±8}{-2} when ± is plus. Add 12 to 8.

$$q=\frac{20}{-2}$$

Divide 20 by -2.

$$q=-10$$

Now solve the equation q=\frac{12±8}{-2} when ± is minus. Subtract 8 from 12.

$$q=\frac{4}{-2}$$

Divide 4 by -2.

$$q=-2$$

The equation is now solved.

$$q=-10 q=-2$$

Steps for Completing the Square

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

$$-12q-q^{2}=20$$

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.

$$-q^{2}-12q=20$$

Divide both sides by -1.

$$\frac{-q^{2}-12q}{-1}=\frac{20}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$q^{2}+\left(-\frac{12}{-1}\right)q=\frac{20}{-1}$$

Divide -12 by -1.

$$q^{2}+12q=\frac{20}{-1}$$

Divide 20 by -1.

$$q^{2}+12q=-20$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}+12q+6^{2}=-20+6^{2}$$

Square 6.

$$q^{2}+12q+36=-20+36$$

Add -20 to 36.

$$q^{2}+12q+36=16$$

Factor q^{2}+12q+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q+6\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q+6\right)^{2}}=\sqrt{16}$$

Simplify.

$$q+6=4 q+6=-4$$

Subtract 6 from both sides of the equation.

$$q=-2 q=-10$$