Solve 9\times(x)^(2)=-15\timesx | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$9 \times { x }^{ 2 } =-15 \times x$$

Solve for x

$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Add $15x$ to both sides.

$$9x^{2}+15x=0$$

Factor out $x$.

$$x\left(9x+15\right)=0$$

To find equation solutions, solve $x=0$ and $9x+15=0$.

$$x=0$$ $$x=-\frac{5}{3}$$

Steps Using the Quadratic Formula

Add $15x$ to both sides.

$$9x^{2}+15x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $9$ for $a$, $15$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-15±\sqrt{15^{2}}}{2\times 9}$$

Take the square root of $15^{2}$.

$$x=\frac{-15±15}{2\times 9}$$

Multiply $2$ times $9$.

$$x=\frac{-15±15}{18}$$

Now solve the equation $x=\frac{-15±15}{18}$ when $±$ is plus. Add $-15$ to $15$.

$$x=\frac{0}{18}$$

Divide $0$ by $18$.

$$x=0$$

Now solve the equation $x=\frac{-15±15}{18}$ when $±$ is minus. Subtract $15$ from $-15$.

$$x=-\frac{30}{18}$$

Reduce the fraction $\frac{-30}{18}$ to lowest terms by extracting and canceling out $6$.

$$x=-\frac{5}{3}$$

The equation is now solved.

$$x=0$$ $$x=-\frac{5}{3}$$

Steps for Completing the Square

Add $15x$ to both sides.

$$9x^{2}+15x=0$$

Divide both sides by $9$.

$$\frac{9x^{2}+15x}{9}=\frac{0}{9}$$

Dividing by $9$ undoes the multiplication by $9$.

$$x^{2}+\frac{15}{9}x=\frac{0}{9}$$

Reduce the fraction $\frac{15}{9}$ to lowest terms by extracting and canceling out $3$.

$$x^{2}+\frac{5}{3}x=\frac{0}{9}$$

Divide $0$ by $9$.

$$x^{2}+\frac{5}{3}x=0$$

Divide $\frac{5}{3}$, the coefficient of the $x$ term, by $2$ to get $\frac{5}{6}$. Then add the square of $\frac{5}{6}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=\left(\frac{5}{6}\right)^{2}$$

Square $\frac{5}{6}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{25}{36}$$

Factor $x^{2}+\frac{5}{3}x+\frac{25}{36}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x+\frac{5}{6}=\frac{5}{6}$$ $$x+\frac{5}{6}=-\frac{5}{6}$$

Subtract $\frac{5}{6}$ from both sides of the equation.

$$x=0$$ $$x=-\frac{5}{3}$$