Algebra Examples

$2 x y^{2} + x^{3} + y = 20$

Step 1

Subtract $20$ from both sides of the equation.

$2 x y^{2} + x^{3} + y - 20 = 0$

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 $a = 2 x$ , $b = 1$ , and $c = x^{3} - 20$ into the quadratic formula and solve for $y$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (2 x \cdot (x^{3} - 20))}}{2 (2 x)}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

One to any power is one.

$y = \frac{- 1 \pm \sqrt{1 - 4 \cdot (2 x) \cdot (x^{3} - 20)}}{2 \cdot (2 x)}$

Step 4.1.2

Multiply $- 4$ by $2$ .

$y = \frac{- 1 \pm \sqrt{1 - 8 x \cdot (x^{3} - 20)}}{2 \cdot (2 x)}$

Step 4.1.3

Apply the distributive property.

$y = \frac{- 1 \pm \sqrt{1 - 8 x \cdot x^{3} - 8 x \cdot - 20}}{2 \cdot (2 x)}$

Step 4.1.4

Multiply $x$ by $x^{3}$ by adding the exponents.

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Step 4.1.4.1

Move $x^{3}$ .

$y = \frac{- 1 \pm \sqrt{1 - 8 (x^{3} x) - 8 x \cdot - 20}}{2 \cdot (2 x)}$

Step 4.1.4.2

Multiply $x^{3}$ by $x$ .

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Step 4.1.4.2.1

Raise $x$ to the power of $1$ .

$y = \frac{- 1 \pm \sqrt{1 - 8 (x^{3} x) - 8 x \cdot - 20}}{2 \cdot (2 x)}$

Step 4.1.4.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = \frac{- 1 \pm \sqrt{1 - 8 x^{3 + 1} - 8 x \cdot - 20}}{2 \cdot (2 x)}$

Step 4.1.4.3

Add $3$ and $1$ .

$y = \frac{- 1 \pm \sqrt{1 - 8 x^{4} - 8 x \cdot - 20}}{2 \cdot (2 x)}$

Step 4.1.5

Multiply $- 20$ by $- 8$ .

$y = \frac{- 1 \pm \sqrt{1 - 8 x^{4} + 160 x}}{2 \cdot (2 x)}$

Step 4.1.6

Reorder terms.

$y = \frac{- 1 \pm \sqrt{- 8 x^{4} + 160 x + 1}}{2 \cdot (2 x)}$

Step 4.2

Multiply $2$ by $2$ .

$y = \frac{- 1 \pm \sqrt{- 8 x^{4} + 160 x + 1}}{4 x}$

Step 5

The final answer is the combination of both solutions.

$y = - \frac{1 - \sqrt{- 8 x^{4} + 160 x + 1}}{4 x}$

$y = - \frac{1 + \sqrt{- 8 x^{4} + 160 x + 1}}{4 x}$