Precalculus Examples

$h y^{2} - k y = p$

Step 1

Subtract $p$ from both sides of the equation.

$h y^{2} - k y - p = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = h$ , $b = - k$ , and $c = - p$ into the quadratic formula and solve for $y$ .

$\frac{k \pm \sqrt{{(- k)}^{2} - 4 \cdot (h \cdot (- p))}}{2 h}$

Step 4

Simplify the numerator.

Tap for more steps...

Step 4.1

Apply the product rule to $- k$ .

$y = \frac{k \pm \sqrt{{(- 1)}^{2} k^{2} - 4 \cdot h \cdot (- 1 p)}}{2 h}$

Step 4.2

Raise $- 1$ to the power of $2$ .

$y = \frac{k \pm \sqrt{1 k^{2} - 4 \cdot h \cdot (- 1 p)}}{2 h}$

Step 4.3

Multiply $k^{2}$ by $1$ .

$y = \frac{k \pm \sqrt{k^{2} - 4 \cdot h \cdot (- 1 p)}}{2 h}$

Step 4.4

Multiply $- 1$ by $- 4$ .

$y = \frac{k \pm \sqrt{k^{2} + 4 h p}}{2 h}$

Step 5

The final answer is the combination of both solutions.

$y = \frac{k + \sqrt{k^{2} + 4 h p}}{2 h}$

$y = \frac{k - \sqrt{k^{2} + 4 h p}}{2 h}$