Solve x - 3(x + 2) при x = 5 | Equatix - Math Solver

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Example

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

a⋅(b-2)=3b

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

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Question

$$x - 3 ( x + 2 ) x = 5$$

Solve for x (complex solution)

$x=\frac{-\sqrt{35}i-5}{6}\approx -0.833333333-0.986013297i$
$x=\frac{-5+\sqrt{35}i}{6}\approx -0.833333333+0.986013297i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $5$ from both sides.

$$x-3\left(x+2\right)x-5=0$$

Use the distributive property to multiply $-3$ by $x+2$.

$$x+\left(-3x-6\right)x-5=0$$

Use the distributive property to multiply $-3x-6$ by $x$.

$$x-3x^{2}-6x-5=0$$

Combine $x$ and $-6x$ to get $-5x$.

$$-5x-3x^{2}-5=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-3x^{2}-5x-5=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-3$ for $a$, $-5$ for $b$, and $-5$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}$$

Square $-5$.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}$$

Multiply $-4$ times $-3$.

$$x=\frac{-\left(-5\right)±\sqrt{25+12\left(-5\right)}}{2\left(-3\right)}$$

Multiply $12$ times $-5$.

$$x=\frac{-\left(-5\right)±\sqrt{25-60}}{2\left(-3\right)}$$

Add $25$ to $-60$.

$$x=\frac{-\left(-5\right)±\sqrt{-35}}{2\left(-3\right)}$$

Take the square root of $-35$.

$$x=\frac{-\left(-5\right)±\sqrt{35}i}{2\left(-3\right)}$$

The opposite of $-5$ is $5$.

$$x=\frac{5±\sqrt{35}i}{2\left(-3\right)}$$

Multiply $2$ times $-3$.

$$x=\frac{5±\sqrt{35}i}{-6}$$

Now solve the equation $x=\frac{5±\sqrt{35}i}{-6}$ when $±$ is plus. Add $5$ to $i\sqrt{35}$.

$$x=\frac{5+\sqrt{35}i}{-6}$$

Divide $5+i\sqrt{35}$ by $-6$.

$$x=\frac{-\sqrt{35}i-5}{6}$$

Now solve the equation $x=\frac{5±\sqrt{35}i}{-6}$ when $±$ is minus. Subtract $i\sqrt{35}$ from $5$.

$$x=\frac{-\sqrt{35}i+5}{-6}$$

Divide $5-i\sqrt{35}$ by $-6$.

$$x=\frac{-5+\sqrt{35}i}{6}$$

The equation is now solved.

$$x=\frac{-\sqrt{35}i-5}{6}$$ $$x=\frac{-5+\sqrt{35}i}{6}$$

Steps for Completing the Square

Multiply $-1$ and $3$ to get $-3$.

$$x-3\left(x+2\right)x=5$$

Use the distributive property to multiply $-3$ by $x+2$.

$$x+\left(-3x-6\right)x=5$$

Use the distributive property to multiply $-3x-6$ by $x$.

$$x-3x^{2}-6x=5$$

Combine $x$ and $-6x$ to get $-5x$.

$$-5x-3x^{2}=5$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-3x^{2}-5x=5$$

Divide both sides by $-3$.

$$\frac{-3x^{2}-5x}{-3}=\frac{5}{-3}$$

Dividing by $-3$ undoes the multiplication by $-3$.

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

Divide $-5$ by $-3$.

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

Divide $5$ by $-3$.

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

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.

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

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

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

Add $-\frac{5}{3}$ to $\frac{25}{36}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=-\frac{35}{36}$$

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 $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{5}{6}\right)^{2}=-\frac{35}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{-\frac{35}{36}}$$

Simplify.

$$x+\frac{5}{6}=\frac{\sqrt{35}i}{6}$$ $$x+\frac{5}{6}=-\frac{\sqrt{35}i}{6}$$

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

$$x=\frac{-5+\sqrt{35}i}{6}$$ $$x=\frac{-\sqrt{35}i-5}{6}$$