Solve Using the Quadratic Formula 5/3k^2-k-6/5=0

$\frac{5}{3} k^{2} - k - \frac{6}{5} = 0$

Step 1

Simplify the left side.

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

$\frac{5 k^{2}}{3} - k - \frac{6}{5} = 0$

Step 2

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

Apply the distributive property.

$15 (\frac{5 k^{2}}{3}) + 15 (- k) + 15 (- \frac{6}{5}) = 0$

Simplify.

Cancel the common factor of $3$ .

Factor $3$ out of $15$ .

$3 (5) (\frac{5 k^{2}}{3}) + 15 (- k) + 15 (- \frac{6}{5}) = 0$

Cancel the common factor.

$3 \cdot (5 (\frac{5 k^{2}}{3})) + 15 (- k) + 15 (- \frac{6}{5}) = 0$

Rewrite the expression.

$5 (5 k^{2}) + 15 (- k) + 15 (- \frac{6}{5}) = 0$

Multiply $5$ by $5$ .

$25 k^{2} + 15 (- k) + 15 (- \frac{6}{5}) = 0$

Multiply $- 1$ by $15$ .

$25 k^{2} - 15 k + 15 (- \frac{6}{5}) = 0$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{6}{5}$ into the numerator.

$25 k^{2} - 15 k + 15 (\frac{- 6}{5}) = 0$

Factor $5$ out of $15$ .

$25 k^{2} - 15 k + 5 (3) (\frac{- 6}{5}) = 0$

Cancel the common factor.

$25 k^{2} - 15 k + 5 \cdot (3 (\frac{- 6}{5})) = 0$

Rewrite the expression.

$25 k^{2} - 15 k + 3 \cdot - 6 = 0$

Multiply $3$ by $- 6$ .

$25 k^{2} - 15 k - 18 = 0$

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 $a = 25$ , $b = - 15$ , and $c = - 18$ into the quadratic formula and solve for $k$ .

$\frac{15 \pm \sqrt{{(- 15)}^{2} - 4 \cdot (25 \cdot - 18)}}{2 \cdot 25}$

Step 5

Simplify.

Simplify the numerator.

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

$k = \frac{15 \pm \sqrt{225 - 4 \cdot 25 \cdot - 18}}{2 \cdot 25}$

Multiply $- 4 \cdot 25 \cdot - 18$ .

Multiply $- 4$ by $25$ .

$k = \frac{15 \pm \sqrt{225 - 100 \cdot - 18}}{2 \cdot 25}$

Multiply $- 100$ by $- 18$ .

$k = \frac{15 \pm \sqrt{225 + 1800}}{2 \cdot 25}$

Add $225$ and $1800$ .

$k = \frac{15 \pm \sqrt{2025}}{2 \cdot 25}$

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

$k = \frac{15 \pm \sqrt{45^{2}}}{2 \cdot 25}$

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

$k = \frac{15 \pm 45}{2 \cdot 25}$

Multiply $2$ by $25$ .

$k = \frac{15 \pm 45}{50}$

Simplify $\frac{15 \pm 45}{50}$ .

$k = \frac{3 \pm 9}{10}$

Step 6

The final answer is the combination of both solutions.

$k = \frac{6}{5}, - \frac{3}{5}$

Solve Using the Quadratic Formula 5/3k^2-k-6/5=0

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