Basic Math Examples

16 y^{2} + z y + 25 = 0

$16 y^{2} + z y + 25 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

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

Step 2

Substitute the values

a = 16

$a = 16$ ,

b = z

$b = z$ , and

c = 25

$c = 25$ into the quadratic formula and solve for

y

$y$ .

\frac{- z \pm \sqrt{z^{2} - 4 \cdot (16 \cdot 25)}}{2 \cdot 16}

$\frac{- z \pm \sqrt{z^{2} - 4 \cdot (16 \cdot 25)}}{2 \cdot 16}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Rewrite

4 \cdot 16 \cdot 25

$4 \cdot 16 \cdot 25$ as

{(2 \cdot 4 \cdot 5)}^{2}

${(2 \cdot 4 \cdot 5)}^{2}$ .

y = \frac{- z \pm \sqrt{z^{2} - {(2 \cdot 4 \cdot 5)}^{2}}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{z^{2} - {(2 \cdot 4 \cdot 5)}^{2}}}{2 \cdot 16}$

Step 3.1.2

Since both terms are perfect squares, factor using the difference of squares formula,

a^{2} - b^{2} = (a + b) (a - b)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = z

$a = z$ and

b = 2 \cdot 4 \cdot 5

$b = 2 \cdot 4 \cdot 5$ .

y = \frac{- z \pm \sqrt{(z + 2 \cdot 4 \cdot 5) (z - (2 \cdot 4 \cdot 5))}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 2 \cdot 4 \cdot 5) (z - (2 \cdot 4 \cdot 5))}}{2 \cdot 16}$

Step 3.1.3

Simplify.

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Step 3.1.3.1

Multiply

2 \cdot 4 \cdot 5

$2 \cdot 4 \cdot 5$ .

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Step 3.1.3.1.1

Multiply

2

$2$ by

4

$4$ .

y = \frac{- z \pm \sqrt{(z + 8 \cdot 5) (z - (2 \cdot 4 \cdot 5))}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 8 \cdot 5) (z - (2 \cdot 4 \cdot 5))}}{2 \cdot 16}$

Step 3.1.3.1.2

Multiply

8

$8$ by

5

$5$ .

y = \frac{- z \pm \sqrt{(z + 40) (z - (2 \cdot 4 \cdot 5))}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 40) (z - (2 \cdot 4 \cdot 5))}}{2 \cdot 16}$

y = \frac{- z \pm \sqrt{(z + 40) (z - (2 \cdot 4 \cdot 5))}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 40) (z - (2 \cdot 4 \cdot 5))}}{2 \cdot 16}$

Step 3.1.3.2

Multiply

- (2 \cdot 4 \cdot 5)

$- (2 \cdot 4 \cdot 5)$ .

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Step 3.1.3.2.1

Multiply

2

$2$ by

4

$4$ .

y = \frac{- z \pm \sqrt{(z + 40) (z - (8 \cdot 5))}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 40) (z - (8 \cdot 5))}}{2 \cdot 16}$

Step 3.1.3.2.2

Multiply

8

$8$ by

5

$5$ .

y = \frac{- z \pm \sqrt{(z + 40) (z - 1 \cdot 40)}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 40) (z - 1 \cdot 40)}}{2 \cdot 16}$

Step 3.1.3.2.3

Multiply

- 1

$- 1$ by

40

$40$ .

y = \frac{- z \pm \sqrt{(z + 40) (z - 40)}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 40) (z - 40)}}{2 \cdot 16}$

y = \frac{- z \pm \sqrt{(z + 40) (z - 40)}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 40) (z - 40)}}{2 \cdot 16}$

y = \frac{- z \pm \sqrt{(z + 40) (z - 40)}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 40) (z - 40)}}{2 \cdot 16}$

y = \frac{- z \pm \sqrt{(z + 40) (z - 40)}}{2 \cdot 16}

$y = \frac{- z \pm \sqrt{(z + 40) (z - 40)}}{2 \cdot 16}$

Step 3.2

Multiply

$2$ by

$16$ .

$y = \frac{- z \pm \sqrt{(z + 40) (z - 40)}}{32}$

Step 4

The final answer is the combination of both solutions.

$y = - \frac{z - \sqrt{(z + 40) (z - 40)}}{32}$

$y = - \frac{z + \sqrt{(z + 40) (z - 40)}}{32}$