How to solve 15(x)^(2)=-10x

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 15(x)^(2)=-10x | Equatix

Question

$$15 { x }^{ 2 } =-10x$$

Solve for x

$x=-\frac{2}{3}\approx -0.666666667$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Add $10x$ to both sides.

$$15x^{2}+10x=0$$

Factor out $x$.

$$x\left(15x+10\right)=0$$

To find equation solutions, solve $x=0$ and $15x+10=0$.

$$x=0$$ $$x=-\frac{2}{3}$$

Steps Using the Quadratic Formula

Add $10x$ to both sides.

$$15x^{2}+10x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $15$ for $a$, $10$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-10±\sqrt{10^{2}}}{2\times 15}$$

Take the square root of $10^{2}$.

$$x=\frac{-10±10}{2\times 15}$$

Multiply $2$ times $15$.

$$x=\frac{-10±10}{30}$$

Now solve the equation $x=\frac{-10±10}{30}$ when $±$ is plus. Add $-10$ to $10$.

$$x=\frac{0}{30}$$

Divide $0$ by $30$.

$$x=0$$

Now solve the equation $x=\frac{-10±10}{30}$ when $±$ is minus. Subtract $10$ from $-10$.

$$x=-\frac{20}{30}$$

Reduce the fraction $\frac{-20}{30}$ to lowest terms by extracting and canceling out $10$.

$$x=-\frac{2}{3}$$

The equation is now solved.

$$x=0$$ $$x=-\frac{2}{3}$$

Steps for Completing the Square

Add $10x$ to both sides.

$$15x^{2}+10x=0$$

Divide both sides by $15$.

$$\frac{15x^{2}+10x}{15}=\frac{0}{15}$$

Dividing by $15$ undoes the multiplication by $15$.

$$x^{2}+\frac{10}{15}x=\frac{0}{15}$$

Reduce the fraction $\frac{10}{15}$ to lowest terms by extracting and canceling out $5$.

$$x^{2}+\frac{2}{3}x=\frac{0}{15}$$

Divide $0$ by $15$.

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

Divide $\frac{2}{3}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{3}$. Then add the square of $\frac{1}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\left(\frac{1}{3}\right)^{2}$$

Square $\frac{1}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{1}{9}$$

Factor $x^{2}+\frac{2}{3}x+\frac{1}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{1}{3}$$ $$x+\frac{1}{3}=-\frac{1}{3}$$

Subtract $\frac{1}{3}$ from both sides of the equation.

$$x=0$$ $$x=-\frac{2}{3}$$

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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