Solve 14q-q^2

$$14 q - q ^ { 2 }$$

Factor

$$q\left(14-q\right)$$

Solution Steps

Factor out q.

$$q\left(14-q\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-q^{2}+14q=0$$

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.

$$q=\frac{-14±\sqrt{14^{2}}}{2\left(-1\right)}$$

Take the square root of 14^{2}.

$$q=\frac{-14±14}{2\left(-1\right)}$$

Multiply 2 times -1.

$$q=\frac{-14±14}{-2}$$

Now solve the equation q=\frac{-14±14}{-2} when ± is plus. Add -14 to 14.

$$q=\frac{0}{-2}$$

Divide 0 by -2.

$$q=0$$

Now solve the equation q=\frac{-14±14}{-2} when ± is minus. Subtract 14 from -14.

$$q=-\frac{28}{-2}$$

Divide -28 by -2.

$$q=14$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and 14 for x_{2}.

$$-q^{2}+14q=-q\left(q-14\right)$$

Evaluate

$$q\left(14-q\right)$$