Basic Math Examples

$- 2 y^{2} - 5 = 4 y$

Step 1

Subtract $4 y$ from both sides of the equation.

$- 2 y^{2} - 5 - 4 y = 0$

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 $a = - 2$ , $b = - 4$ , and $c = - 5$ into the quadratic formula and solve for $y$ .

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 4$ to the power of $2$ .

$y = \frac{4 \pm \sqrt{16 - 4 \cdot - 2 \cdot - 5}}{2 \cdot - 2}$

Step 4.1.2

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

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Step 4.1.2.1

Multiply $- 4$ by $- 2$ .

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

Step 4.1.2.2

Multiply $8$ by $- 5$ .

$y = \frac{4 \pm \sqrt{16 - 40}}{2 \cdot - 2}$

Step 4.1.3

Subtract $40$ from $16$ .

$y = \frac{4 \pm \sqrt{- 24}}{2 \cdot - 2}$

Step 4.1.4

Rewrite $- 24$ as $- 1 (24)$ .

$y = \frac{4 \pm \sqrt{- 1 \cdot 24}}{2 \cdot - 2}$

Step 4.1.5

Rewrite $\sqrt{- 1 (24)}$ as $\sqrt{- 1} \cdot \sqrt{24}$ .

$y = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{24}}{2 \cdot - 2}$

Step 4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{4 \pm i \cdot \sqrt{24}}{2 \cdot - 2}$

Step 4.1.7

Rewrite $24$ as $2^{2} \cdot 6$ .

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Step 4.1.7.1

Factor $4$ out of $24$ .

$y = \frac{4 \pm i \cdot \sqrt{4 (6)}}{2 \cdot - 2}$

Step 4.1.7.2

Rewrite $4$ as $2^{2}$ .

$y = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 6}}{2 \cdot - 2}$

Step 4.1.8

Pull terms out from under the radical.

$y = \frac{4 \pm i \cdot (2 \sqrt{6})}{2 \cdot - 2}$

Step 4.1.9

Move $2$ to the left of $i$ .

$y = \frac{4 \pm 2 i \sqrt{6}}{2 \cdot - 2}$

Step 4.2

Multiply $2$ by $- 2$ .

$y = \frac{4 \pm 2 i \sqrt{6}}{- 4}$

Step 4.3

Simplify $\frac{4 \pm 2 i \sqrt{6}}{- 4}$ .

$y = \frac{2 \pm i \sqrt{6}}{- 2}$

Step 4.4

Move the negative in front of the fraction.

$y = - \frac{2 \pm i \sqrt{6}}{2}$

Step 5

The final answer is the combination of both solutions.

$y = - \frac{2 + i \sqrt{6}}{2}, - \frac{2 - i \sqrt{6}}{2}$