Solve 339=-q^2-2q+534

$$339 = - q ^ { 2 } - 2 q + 534$$

Solve for q

$$q=-15$$

$$q=13$$

Steps Using Factoring By Grouping

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

$$-q^{2}-2q+534=339$$

Subtract 339 from both sides.

$$-q^{2}-2q+534-339=0$$

Subtract 339 from 534 to get 195.

$$-q^{2}-2q+195=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -q^{2}+aq+bq+195. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-195=-195$$

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 -195.

$$1,-195 3,-65 5,-39 13,-15$$

Calculate the sum for each pair.

$$1-195=-194 3-65=-62 5-39=-34 13-15=-2$$

The solution is the pair that gives sum -2.

$$a=13 b=-15$$

Rewrite -q^{2}-2q+195 as \left(-q^{2}+13q\right)+\left(-15q+195\right).

$$\left(-q^{2}+13q\right)+\left(-15q+195\right)$$

Factor out q in the first and 15 in the second group.

$$q\left(-q+13\right)+15\left(-q+13\right)$$

Factor out common term -q+13 by using distributive property.

$$\left(-q+13\right)\left(q+15\right)$$

To find equation solutions, solve -q+13=0 and q+15=0.

$$q=13 q=-15$$

Steps Using the Quadratic Formula

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

$$-q^{2}-2q+534=339$$

Subtract 339 from both sides.

$$-q^{2}-2q+534-339=0$$

Subtract 339 from 534 to get 195.

$$-q^{2}-2q+195=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 195 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 195}}{2\left(-1\right)}$$

Square -2.

$$q=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 195}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$q=\frac{-\left(-2\right)±\sqrt{4+4\times 195}}{2\left(-1\right)}$$

Multiply 4 times 195.

$$q=\frac{-\left(-2\right)±\sqrt{4+780}}{2\left(-1\right)}$$

Add 4 to 780.

$$q=\frac{-\left(-2\right)±\sqrt{784}}{2\left(-1\right)}$$

Take the square root of 784.

$$q=\frac{-\left(-2\right)±28}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$q=\frac{2±28}{2\left(-1\right)}$$

Multiply 2 times -1.

$$q=\frac{2±28}{-2}$$

Now solve the equation q=\frac{2±28}{-2} when ± is plus. Add 2 to 28.

$$q=\frac{30}{-2}$$

Divide 30 by -2.

$$q=-15$$

Now solve the equation q=\frac{2±28}{-2} when ± is minus. Subtract 28 from 2.

$$q=-\frac{26}{-2}$$

Divide -26 by -2.

$$q=13$$

The equation is now solved.

$$q=-15 q=13$$

Steps for Completing the Square

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

$$-q^{2}-2q+534=339$$

Subtract 534 from both sides.

$$-q^{2}-2q=339-534$$

Subtract 534 from 339 to get -195.

$$-q^{2}-2q=-195$$

Divide both sides by -1.

$$\frac{-q^{2}-2q}{-1}=-\frac{195}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$q^{2}+\left(-\frac{2}{-1}\right)q=-\frac{195}{-1}$$

Divide -2 by -1.

$$q^{2}+2q=-\frac{195}{-1}$$

Divide -195 by -1.

$$q^{2}+2q=195$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}+2q+1^{2}=195+1^{2}$$

Square 1.

$$q^{2}+2q+1=195+1$$

Add 195 to 1.

$$q^{2}+2q+1=196$$

Factor q^{2}+2q+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(q+1\right)^{2}=196$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q+1\right)^{2}}=\sqrt{196}$$

Simplify.

$$q+1=14 q+1=-14$$

Subtract 1 from both sides of the equation.

$$q=13 q=-15$$