Solve for y -75=y^3+50

$- 75 = y^{3} + 50$

Rewrite the equation as $y^{3} + 50 = - 75$ .

$y^{3} + 50 = - 75$

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

Subtract $50$ from both sides of the equation.

$y^{3} = - 75 - 50$

Subtract $50$ from $- 75$ .

$y^{3} = - 125$

Add $125$ to both sides of the equation.

$y^{3} + 125 = 0$

Factor the left side of the equation.

Rewrite $125$ as $5^{3}$ .

$y^{3} + 5^{3} = 0$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = y$ and $b = 5$ .

$(y + 5) (y^{2} - y \cdot 5 + 5^{2}) = 0$

Simplify.

Multiply $5$ by $- 1$ .

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

Raise $5$ to the power of $2$ .

$(y + 5) (y^{2} - 5 y + 25) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y + 5 = 0$

$y^{2} - 5 y + 25 = 0$

Set $y + 5$ equal to $0$ and solve for $y$ .

Set $y + 5$ equal to $0$ .

$y + 5 = 0$

Subtract $5$ from both sides of the equation.

$y = - 5$

Set $y^{2} - 5 y + 25$ equal to $0$ and solve for $y$ .

Set $y^{2} - 5 y + 25$ equal to $0$ .

$y^{2} - 5 y + 25 = 0$

Solve $y^{2} - 5 y + 25 = 0$ for $y$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 5$ , and $c = 25$ into the quadratic formula and solve for $y$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot 25)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot 25}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 25}}{2 \cdot 1}$

Multiply $- 4$ by $25$ .

$y = \frac{5 \pm \sqrt{25 - 100}}{2 \cdot 1}$

Subtract $100$ from $25$ .

$y = \frac{5 \pm \sqrt{- 75}}{2 \cdot 1}$

Rewrite $- 75$ as $- 1 (75)$ .

$y = \frac{5 \pm \sqrt{- 1 \cdot 75}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (75)}$ as $\sqrt{- 1} \cdot \sqrt{75}$ .

$y = \frac{5 \pm \sqrt{- 1} \cdot \sqrt{75}}{2 \cdot 1}$

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

$y = \frac{5 \pm i \cdot \sqrt{75}}{2 \cdot 1}$

Rewrite $75$ as $5^{2} \cdot 3$ .

Factor $25$ out of $75$ .

$y = \frac{5 \pm i \cdot \sqrt{25 (3)}}{2 \cdot 1}$

Rewrite $25$ as $5^{2}$ .

$y = \frac{5 \pm i \cdot \sqrt{5^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{5 \pm i \cdot (5 \sqrt{3})}{2 \cdot 1}$

Move $5$ to the left of $i$ .

$y = \frac{5 \pm 5 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{5 \pm 5 i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$y = \frac{5 + 5 i \sqrt{3}}{2}, \frac{5 - 5 i \sqrt{3}}{2}$

The final solution is all the values that make $(y + 5) (y^{2} - 5 y + 25) = 0$ true.

$y = - 5, \frac{5 + 5 i \sqrt{3}}{2}, \frac{5 - 5 i \sqrt{3}}{2}$

Solve for y -75=y^3+50

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