Solve for w w(w-10)+40=0

$w (w - 10) + 40 = 0$

Step 1

Simplify each term.

Apply the distributive property.

$w \cdot w + w \cdot - 10 + 40 = 0$

Multiply $w$ by $w$ .

$w^{2} + w \cdot - 10 + 40 = 0$

Move $- 10$ to the left of $w$ .

$w^{2} - 10 w + 40 = 0$

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 $a = 1$ , $b = - 10$ , and $c = 40$ into the quadratic formula and solve for $w$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (1 \cdot 40)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 10$ to the power of $2$ .

$w = \frac{10 \pm \sqrt{100 - 4 \cdot 1 \cdot 40}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 40$ .

Multiply $- 4$ by $1$ .

$w = \frac{10 \pm \sqrt{100 - 4 \cdot 40}}{2 \cdot 1}$

Multiply $- 4$ by $40$ .

$w = \frac{10 \pm \sqrt{100 - 160}}{2 \cdot 1}$

Subtract $160$ from $100$ .

$w = \frac{10 \pm \sqrt{- 60}}{2 \cdot 1}$

Rewrite $- 60$ as $- 1 (60)$ .

$w = \frac{10 \pm \sqrt{- 1 \cdot 60}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (60)}$ as $\sqrt{- 1} \cdot \sqrt{60}$ .

$w = \frac{10 \pm \sqrt{- 1} \cdot \sqrt{60}}{2 \cdot 1}$

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

$w = \frac{10 \pm i \cdot \sqrt{60}}{2 \cdot 1}$

Rewrite $60$ as $2^{2} \cdot 15$ .

Factor $4$ out of $60$ .

$w = \frac{10 \pm i \cdot \sqrt{4 (15)}}{2 \cdot 1}$

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

$w = \frac{10 \pm i \cdot \sqrt{2^{2} \cdot 15}}{2 \cdot 1}$

Pull terms out from under the radical.

$w = \frac{10 \pm i \cdot (2 \sqrt{15})}{2 \cdot 1}$

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

$w = \frac{10 \pm 2 i \sqrt{15}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$w = \frac{10 \pm 2 i \sqrt{15}}{2}$

Simplify $\frac{10 \pm 2 i \sqrt{15}}{2}$ .

$w = 5 \pm i \sqrt{15}$

Step 5

The final answer is the combination of both solutions.

$w = 5 + i \sqrt{15}, 5 - i \sqrt{15}$

Solve for w w(w-10)+40=0

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