Finite Math Examples

$2 y^{2} + (2 k - 3) y - 5 = 0$

Step 1

Apply the distributive property.

$2 y^{2} + 2 k y - 3 y - 5 = 0$

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 $a = 2$ , $b = 2 k - 3$ , and $c = - 5$ into the quadratic formula and solve for $y$ .

$\frac{- (2 k - 3) \pm \sqrt{{(2 k - 3)}^{2} - 4 \cdot (2 \cdot - 5)}}{2 \cdot 2}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Apply the distributive property.

$y = \frac{- (2 k) + 3 \pm \sqrt{{(2 k - 3)}^{2} - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.2

Multiply $2$ by $- 1$ .

$y = \frac{- 2 k + 3 \pm \sqrt{{(2 k - 3)}^{2} - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.3

Multiply $- 1$ by $- 3$ .

$y = \frac{- 2 k + 3 \pm \sqrt{{(2 k - 3)}^{2} - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.4

Rewrite ${(2 k - 3)}^{2}$ as $(2 k - 3) (2 k - 3)$ .

$y = \frac{- 2 k + 3 \pm \sqrt{(2 k - 3) (2 k - 3) - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.5

Expand $(2 k - 3) (2 k - 3)$ using the FOIL Method.

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Step 4.1.5.1

Apply the distributive property.

$y = \frac{- 2 k + 3 \pm \sqrt{2 k (2 k - 3) - 3 (2 k - 3) - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.5.2

Apply the distributive property.

$y = \frac{- 2 k + 3 \pm \sqrt{2 k (2 k) + 2 k \cdot - 3 - 3 (2 k - 3) - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.5.3

Apply the distributive property.

$y = \frac{- 2 k + 3 \pm \sqrt{2 k (2 k) + 2 k \cdot - 3 - 3 (2 k) - 3 \cdot - 3 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.6

Simplify and combine like terms.

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Step 4.1.6.1

Simplify each term.

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Step 4.1.6.1.1

Rewrite using the commutative property of multiplication.

$y = \frac{- 2 k + 3 \pm \sqrt{2 \cdot (2 k \cdot k) + 2 k \cdot - 3 - 3 (2 k) - 3 \cdot - 3 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.6.1.2

Multiply $k$ by $k$ by adding the exponents.

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Step 4.1.6.1.2.1

Move $k$ .

$y = \frac{- 2 k + 3 \pm \sqrt{2 \cdot (2 (k \cdot k)) + 2 k \cdot - 3 - 3 (2 k) - 3 \cdot - 3 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.6.1.2.2

Multiply $k$ by $k$ .

$y = \frac{- 2 k + 3 \pm \sqrt{2 \cdot (2 k^{2}) + 2 k \cdot - 3 - 3 (2 k) - 3 \cdot - 3 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.6.1.3

Multiply $2$ by $2$ .

$y = \frac{- 2 k + 3 \pm \sqrt{4 k^{2} + 2 k \cdot - 3 - 3 (2 k) - 3 \cdot - 3 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.6.1.4

Multiply $- 3$ by $2$ .

$y = \frac{- 2 k + 3 \pm \sqrt{4 k^{2} - 6 k - 3 (2 k) - 3 \cdot - 3 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.6.1.5

Multiply $2$ by $- 3$ .

$y = \frac{- 2 k + 3 \pm \sqrt{4 k^{2} - 6 k - 6 k - 3 \cdot - 3 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.6.1.6

Multiply $- 3$ by $- 3$ .

$y = \frac{- 2 k + 3 \pm \sqrt{4 k^{2} - 6 k - 6 k + 9 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.6.2

Subtract $6 k$ from $- 6 k$ .

$y = \frac{- 2 k + 3 \pm \sqrt{4 k^{2} - 12 k + 9 - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

Step 4.1.7

Multiply $- 4 \cdot 2 \cdot - 5$ .

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Step 4.1.7.1

Multiply $- 4$ by $2$ .

$y = \frac{- 2 k + 3 \pm \sqrt{4 k^{2} - 12 k + 9 - 8 \cdot - 5}}{2 \cdot 2}$

Step 4.1.7.2

Multiply $- 8$ by $- 5$ .

$y = \frac{- 2 k + 3 \pm \sqrt{4 k^{2} - 12 k + 9 + 40}}{2 \cdot 2}$

Step 4.1.8

Add $9$ and $40$ .

$y = \frac{- 2 k + 3 \pm \sqrt{4 k^{2} - 12 k + 49}}{2 \cdot 2}$

Step 4.2

Multiply $2$ by $2$ .

$y = \frac{- 2 k + 3 \pm \sqrt{4 k^{2} - 12 k + 49}}{4}$

Step 5

The final answer is the combination of both solutions.

$y = - \frac{2 k - 3 - \sqrt{4 k^{2} - 12 k + 49}}{4}$

$y = - \frac{2 k - 3 + \sqrt{4 k^{2} - 12 k + 49}}{4}$