How to solve 6x ^ 2 = - 3x

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 6x ^ 2 = - 3x | Equatix

Question

$$6x^{2}=-3x$$

Solve for x

$x=-\frac{1}{2}=-0.5$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Add $3x$ to both sides.

$$6x^{2}+3x=0$$

Factor out $x$.

$$x\left(6x+3\right)=0$$

To find equation solutions, solve $x=0$ and $6x+3=0$.

$$x=0$$ $$x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Add $3x$ to both sides.

$$6x^{2}+3x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $6$ for $a$, $3$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-3±\sqrt{3^{2}}}{2\times 6}$$

Take the square root of $3^{2}$.

$$x=\frac{-3±3}{2\times 6}$$

Multiply $2$ times $6$.

$$x=\frac{-3±3}{12}$$

Now solve the equation $x=\frac{-3±3}{12}$ when $±$ is plus. Add $-3$ to $3$.

$$x=\frac{0}{12}$$

Divide $0$ by $12$.

$$x=0$$

Now solve the equation $x=\frac{-3±3}{12}$ when $±$ is minus. Subtract $3$ from $-3$.

$$x=-\frac{6}{12}$$

Reduce the fraction $\frac{-6}{12}$ to lowest terms by extracting and canceling out $6$.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=0$$ $$x=-\frac{1}{2}$$

Steps for Completing the Square

Add $3x$ to both sides.

$$6x^{2}+3x=0$$

Divide both sides by $6$.

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

Dividing by $6$ undoes the multiplication by $6$.

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

Reduce the fraction $\frac{3}{6}$ to lowest terms by extracting and canceling out $3$.

$$x^{2}+\frac{1}{2}x=\frac{0}{6}$$

Divide $0$ by $6$.

$$x^{2}+\frac{1}{2}x=0$$

Divide $\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{4}$. Then add the square of $\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\left(\frac{1}{4}\right)^{2}$$

Square $\frac{1}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Factor $x^{2}+\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{1}{4}$$ $$x+\frac{1}{4}=-\frac{1}{4}$$

Subtract $\frac{1}{4}$ from both sides of the equation.

$$x=0$$ $$x=-\frac{1}{2}$$

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Example

Question

Solve for x

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square