Precalculus Examples

$5 x^{2} + 7 x - h = 0$

Step 1

Move all terms not containing $h$ to the right side of the equation.

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Step 1.1

Subtract $5 x^{2}$ from both sides of the equation.

$7 x - h = - 5 x^{2}$

Step 1.2

Subtract $7 x$ from both sides of the equation.

$- h = - 5 x^{2} - 7 x$

Step 2

Divide each term in $- h = - 5 x^{2} - 7 x$ by $- 1$ and simplify.

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Step 2.1

Divide each term in $- h = - 5 x^{2} - 7 x$ by $- 1$ .

$\frac{- h}{- 1} = \frac{- 5 x^{2}}{- 1} + \frac{- 7 x}{- 1}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Dividing two negative values results in a positive value.

$\frac{h}{1} = \frac{- 5 x^{2}}{- 1} + \frac{- 7 x}{- 1}$

Step 2.2.2

Divide $h$ by $1$ .

$h = \frac{- 5 x^{2}}{- 1} + \frac{- 7 x}{- 1}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Move the negative one from the denominator of $\frac{- 5 x^{2}}{- 1}$ .

$h = - 1 \cdot (- 5 x^{2}) + \frac{- 7 x}{- 1}$

Step 2.3.1.2

Rewrite $- 1 \cdot (- 5 x^{2})$ as $- (- 5 x^{2})$ .

$h = - (- 5 x^{2}) + \frac{- 7 x}{- 1}$

Step 2.3.1.3

Multiply $- 5$ by $- 1$ .

$h = 5 x^{2} + \frac{- 7 x}{- 1}$

Step 2.3.1.4

Move the negative one from the denominator of $\frac{- 7 x}{- 1}$ .

$h = 5 x^{2} - 1 \cdot (- 7 x)$

Step 2.3.1.5

Rewrite $- 1 \cdot (- 7 x)$ as $- (- 7 x)$ .

$h = 5 x^{2} - (- 7 x)$

Step 2.3.1.6

Multiply $- 7$ by $- 1$ .

$h = 5 x^{2} + 7 x$