Solve Using the Quadratic Formula 3/5k^2-k-10/3=0

$\frac{3}{5} k^{2} - k - \frac{10}{3} = 0$

Step 1

Simplify the left side.

Combine $\frac{3}{5}$ and $k^{2}$ .

$\frac{3 k^{2}}{5} - k - \frac{10}{3} = 0$

Step 2

Multiply through by the least common denominator $15$ , then simplify.

Apply the distributive property.

$15 (\frac{3 k^{2}}{5}) + 15 (- k) + 15 (- \frac{10}{3}) = 0$

Simplify.

Cancel the common factor of $5$ .

Factor $5$ out of $15$ .

$5 (3) (\frac{3 k^{2}}{5}) + 15 (- k) + 15 (- \frac{10}{3}) = 0$

Cancel the common factor.

$5 \cdot (3 (\frac{3 k^{2}}{5})) + 15 (- k) + 15 (- \frac{10}{3}) = 0$

Rewrite the expression.

$3 (3 k^{2}) + 15 (- k) + 15 (- \frac{10}{3}) = 0$

Multiply $3$ by $3$ .

$9 k^{2} + 15 (- k) + 15 (- \frac{10}{3}) = 0$

Multiply $- 1$ by $15$ .

$9 k^{2} - 15 k + 15 (- \frac{10}{3}) = 0$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{10}{3}$ into the numerator.

$9 k^{2} - 15 k + 15 (\frac{- 10}{3}) = 0$

Factor $3$ out of $15$ .

$9 k^{2} - 15 k + 3 (5) (\frac{- 10}{3}) = 0$

Cancel the common factor.

$9 k^{2} - 15 k + 3 \cdot (5 (\frac{- 10}{3})) = 0$

Rewrite the expression.

$9 k^{2} - 15 k + 5 \cdot - 10 = 0$

Multiply $5$ by $- 10$ .

$9 k^{2} - 15 k - 50 = 0$

Step 3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4

Substitute the values $a = 9$ , $b = - 15$ , and $c = - 50$ into the quadratic formula and solve for $k$ .

$\frac{15 \pm \sqrt{{(- 15)}^{2} - 4 \cdot (9 \cdot - 50)}}{2 \cdot 9}$

Step 5

Simplify.

Simplify the numerator.

Raise $- 15$ to the power of $2$ .

$k = \frac{15 \pm \sqrt{225 - 4 \cdot 9 \cdot - 50}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 50$ .

Multiply $- 4$ by $9$ .

$k = \frac{15 \pm \sqrt{225 - 36 \cdot - 50}}{2 \cdot 9}$

Multiply $- 36$ by $- 50$ .

$k = \frac{15 \pm \sqrt{225 + 1800}}{2 \cdot 9}$

Add $225$ and $1800$ .

$k = \frac{15 \pm \sqrt{2025}}{2 \cdot 9}$

Rewrite $2025$ as $45^{2}$ .

$k = \frac{15 \pm \sqrt{45^{2}}}{2 \cdot 9}$

Pull terms out from under the radical, assuming positive real numbers.

$k = \frac{15 \pm 45}{2 \cdot 9}$

Multiply $2$ by $9$ .

$k = \frac{15 \pm 45}{18}$

Simplify $\frac{15 \pm 45}{18}$ .

$k = \frac{5 \pm 15}{6}$

Step 6

The final answer is the combination of both solutions.

$k = \frac{10}{3}, - \frac{5}{3}$

Solve Using the Quadratic Formula 3/5k^2-k-10/3=0

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