Question

Solve the equation for all real solutions in simplest form.
5, c, squared, minus, 21, c, plus, 11, equals, minus, 4, c
5c
2
−21c+11=
−4c

Answer 1

To solve the equation \( 5c^2 - 21c + 11 = -4c \), we first move all terms to one side of the equation:

\[
5c^2 - 21c + 11 + 4c = 0
\]

This simplifies to:

\[
5c^2 - 17c + 11 = 0
\]

Now we need to solve the quadratic equation \( 5c^2 - 17c + 11 = 0 \). We can use the quadratic formula, given by:

\[
c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 5 \), \( b = -17 \), and \( c = 11 \). First, we calculate the discriminant (\( b^2 - 4ac \)):

\[
b^2 - 4ac = (-17)^2 - 4 \cdot 5 \cdot 11
\]
\[
= 289 - 220
\]
\[
= 69
\]

Now that we have the discriminant, we can find the solutions:

\[
c = \frac{-(-17) \pm \sqrt{69}}{2 \cdot 5}
\]
\[
= \frac{17 \pm \sqrt{69}}{10}
\]

So, the solutions for \( c \) are:

\[
c = \frac{17 + \sqrt{69}}{10} \quad \text{and} \quad c = \frac{17 - \sqrt{69}}{10}
\]

Thus, the complete set of real solutions in simplest form is:

\[
c = \frac{17 + \sqrt{69}}{10}, \quad c = \frac{17 - \sqrt{69}}{10}
\]