Find the Roots (Zeros) 3x^2+30=-18x

$3 x^{2} + 30 = - 18 x$

Step 1

Add $18 x$ to both sides of the equation.

$3 x^{2} + 30 + 18 x = 0$

Step 2

Factor the left side of the equation.

Factor $3$ out of $3 x^{2} + 30 + 18 x$ .

Factor $3$ out of $3 x^{2}$ .

$3 (x^{2}) + 30 + 18 x = 0$

Factor $3$ out of $30$ .

$3 x^{2} + 3 \cdot 10 + 18 x = 0$

Factor $3$ out of $18 x$ .

$3 x^{2} + 3 \cdot 10 + 3 (6 x) = 0$

Factor $3$ out of $3 x^{2} + 3 \cdot 10$ .

$3 (x^{2} + 10) + 3 (6 x) = 0$

Factor $3$ out of $3 (x^{2} + 10) + 3 (6 x)$ .

$3 (x^{2} + 10 + 6 x) = 0$

Reorder terms.

$3 (x^{2} + 6 x + 10) = 0$

Step 3

Divide each term in $3 (x^{2} + 6 x + 10) = 0$ by $3$ and simplify.

Divide each term in $3 (x^{2} + 6 x + 10) = 0$ by $3$ .

$\frac{3 (x^{2} + 6 x + 10)}{3} = \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (x^{2} + 6 x + 10)}{3} = \frac{0}{3}$

Divide $x^{2} + 6 x + 10$ by $1$ .

$x^{2} + 6 x + 10 = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$x^{2} + 6 x + 10 = 0$

Step 4

Use the quadratic formula to find the solutions.

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

Step 5

Substitute the values $a = 1$ , $b = 6$ , and $c = 10$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (1 \cdot 10)}}{2 \cdot 1}$

Step 6

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 10$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 10}}{2 \cdot 1}$

Multiply $- 4$ by $10$ .

$x = \frac{- 6 \pm \sqrt{36 - 40}}{2 \cdot 1}$

Subtract $40$ from $36$ .

$x = \frac{- 6 \pm \sqrt{- 4}}{2 \cdot 1}$

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{- 6 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 1}$

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

$x = \frac{- 6 \pm i \cdot \sqrt{4}}{2 \cdot 1}$

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

$x = \frac{- 6 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 6 \pm i \cdot 2}{2 \cdot 1}$

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

$x = \frac{- 6 \pm 2 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 6 \pm 2 i}{2}$

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

$x = - 3 \pm i$

Step 7

Find the Roots (Zeros) 3x^2+30=-18x

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