Basic Math Examples

$2 h^{2} + y = h y$

Step 1

Subtract $h y$ from both sides of the equation.

$2 h^{2} + y - h y = 0$

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 $a = 2$ , $b = - y$ , and $c = y$ into the quadratic formula and solve for $h$ .

$\frac{y \pm \sqrt{{(- y)}^{2} - 4 \cdot (2 \cdot y)}}{2 \cdot 2}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Apply the product rule to $- y$ .

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

Step 4.1.2

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

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

Step 4.1.3

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

$h = \frac{y \pm \sqrt{y^{2} - 4 \cdot 2 \cdot y}}{2 \cdot 2}$

Step 4.1.4

Multiply $- 4$ by $2$ .

$h = \frac{y \pm \sqrt{y^{2} - 8 \cdot y}}{2 \cdot 2}$

Step 4.1.5

Factor $y$ out of $y^{2} - 8 y$ .

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Step 4.1.5.1

Factor $y$ out of $y^{2}$ .

$h = \frac{y \pm \sqrt{y \cdot y - 8 y}}{2 \cdot 2}$

Step 4.1.5.2

Factor $y$ out of $- 8 y$ .

$h = \frac{y \pm \sqrt{y \cdot y + y \cdot - 8}}{2 \cdot 2}$

Step 4.1.5.3

Factor $y$ out of $y \cdot y + y \cdot - 8$ .

$h = \frac{y \pm \sqrt{y (y - 8)}}{2 \cdot 2}$

Step 4.2

Multiply $2$ by $2$ .

$h = \frac{y \pm \sqrt{y (y - 8)}}{4}$

Step 5

The final answer is the combination of both solutions.

$h = \frac{y + \sqrt{y (y - 8)}}{4}$

$h = \frac{y - \sqrt{y (y - 8)}}{4}$