Question

Please help (quadratic formula)

Answer 1

@Mimi_x3

Answer 2

Can you find a copy of the quadratic formula and post it?

Answer 3

\[-b \pm \sqrt{b^2-4ac}/2a\]

Answer 4

The 2a denominator is below the entire expression before it.

Answer 5

Now simply use 5 for a, 3 for b, and -2 for c in the quadratic formula.
Then simplify the expression.

Answer 6

2/5

Answer 7

\(k = \dfrac{-\color{red}{b} \pm \sqrt{\color{red}{b}^2 - 4\color{green}{a}\color{purple}{c}}}{2\color{green}{a}}\)

\(\color{green}{a = 5}\)
\(\color{red}{b = 3}\)
\(\color{purple}{c = -2}\)

\(k = \dfrac{-\color{red}{3} \pm \sqrt{\color{red}{3}^2 - 4(\color{green}{5})(\color{purple}{-2})}}{2(\color{green}{5})}\)

Answer 8

\(k = \dfrac{-3 \pm \sqrt{9 - (-40)}}{10}\)

\(k = \dfrac{-3 \pm \sqrt{49}}{10}\)

\(k = \dfrac{-3 \pm 7}{10}\)

Did you get to this point?

Answer 9

yes

Answer 10

Good.
Now you need to know what to do with the \(\pm\).

Answer 11

After you simplify the solution as much as possible while you still have the \(\pm\), you need to turn it into two equations, one for the + case, and one for the - case. The two equations are separated by the word "or."

Answer 12

\(k = \dfrac{-3 \pm 7}{10}\)

\(k = \dfrac{-3 + 7}{10}\) or \(k = \dfrac{-3 -7}{10}\)

Now you solve each equation above and always kep the word "or" in between them.

Answer 13

\[\frac{ -3-7 }{ 10 }=-1\]

Answer 14

\[\frac{ -3+7 }{ 10 }=\frac{ 2 }{ 5 }\]

Answer 15

\(k = \dfrac{-3 + 7}{10}\) or \(k = \dfrac{-3 - 7}{10}\)

\(k = \dfrac{4}{10}\) or \(k = \dfrac{-10}{10}\)

\(k = \dfrac{2}{5}\) or \(k = -1\)

Answer 16

You are correct. Those two solutions are correct.