How to solve 5x(x - 6) = 1

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 5x(x - 6) = 1 | Equatix

Question

$$5x(x-6)=1$$

Solve for x

$x=\frac{\sqrt{230}}{5}+3\approx 6.033150178$
$x=-\frac{\sqrt{230}}{5}+3\approx -0.033150178$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $5x$ by $x-6$.

$$5x^{2}-30x=1$$

Subtract $1$ from both sides.

$$5x^{2}-30x-1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $-30$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square $-30$.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 5\left(-1\right)}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-\left(-30\right)±\sqrt{900-20\left(-1\right)}}{2\times 5}$$

Multiply $-20$ times $-1$.

$$x=\frac{-\left(-30\right)±\sqrt{900+20}}{2\times 5}$$

Add $900$ to $20$.

$$x=\frac{-\left(-30\right)±\sqrt{920}}{2\times 5}$$

Take the square root of $920$.

$$x=\frac{-\left(-30\right)±2\sqrt{230}}{2\times 5}$$

The opposite of $-30$ is $30$.

$$x=\frac{30±2\sqrt{230}}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{30±2\sqrt{230}}{10}$$

Now solve the equation $x=\frac{30±2\sqrt{230}}{10}$ when $±$ is plus. Add $30$ to $2\sqrt{230}$.

$$x=\frac{2\sqrt{230}+30}{10}$$

Divide $30+2\sqrt{230}$ by $10$.

$$x=\frac{\sqrt{230}}{5}+3$$

Now solve the equation $x=\frac{30±2\sqrt{230}}{10}$ when $±$ is minus. Subtract $2\sqrt{230}$ from $30$.

$$x=\frac{30-2\sqrt{230}}{10}$$

Divide $30-2\sqrt{230}$ by $10$.

$$x=-\frac{\sqrt{230}}{5}+3$$

The equation is now solved.

$$x=\frac{\sqrt{230}}{5}+3$$ $$x=-\frac{\sqrt{230}}{5}+3$$

Steps for Completing the Square

Use the distributive property to multiply $5x$ by $x-6$.

$$5x^{2}-30x=1$$

Divide both sides by $5$.

$$\frac{5x^{2}-30x}{5}=\frac{1}{5}$$

Dividing by $5$ undoes the multiplication by $5$.

$$x^{2}+\left(-\frac{30}{5}\right)x=\frac{1}{5}$$

Divide $-30$ by $5$.

$$x^{2}-6x=\frac{1}{5}$$

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

$$x^{2}-6x+\left(-3\right)^{2}=\frac{1}{5}+\left(-3\right)^{2}$$

Square $-3$.

$$x^{2}-6x+9=\frac{1}{5}+9$$

Add $\frac{1}{5}$ to $9$.

$$x^{2}-6x+9=\frac{46}{5}$$

Factor $x^{2}-6x+9$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-3\right)^{2}=\frac{46}{5}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{\frac{46}{5}}$$

Simplify.

$$x-3=\frac{\sqrt{230}}{5}$$ $$x-3=-\frac{\sqrt{230}}{5}$$

Add $3$ to both sides of the equation.

$$x=\frac{\sqrt{230}}{5}+3$$ $$x=-\frac{\sqrt{230}}{5}+3$$

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Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square