Solve 2x ^ 2 - 9x - 18 = x | 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

$$2x^{2}-9x-18=x$$

Solve for x

$x = \frac{\sqrt{61} + 5}{2} \approx 6.405124838$
$x=\frac{5-\sqrt{61}}{2}\approx -1.405124838$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $x$ from both sides.

$$2x^{2}-9x-18-x=0$$

Combine $-9x$ and $-x$ to get $-10x$.

$$2x^{2}-10x-18=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-10$ for $b$, and $-18$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Square $-10$.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 2\left(-18\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-10\right)±\sqrt{100-8\left(-18\right)}}{2\times 2}$$

Multiply $-8$ times $-18$.

$$x=\frac{-\left(-10\right)±\sqrt{100+144}}{2\times 2}$$

Add $100$ to $144$.

$$x=\frac{-\left(-10\right)±\sqrt{244}}{2\times 2}$$

Take the square root of $244$.

$$x=\frac{-\left(-10\right)±2\sqrt{61}}{2\times 2}$$

The opposite of $-10$ is $10$.

$$x=\frac{10±2\sqrt{61}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{10±2\sqrt{61}}{4}$$

Now solve the equation $x=\frac{10±2\sqrt{61}}{4}$ when $±$ is plus. Add $10$ to $2\sqrt{61}$.

$$x=\frac{2\sqrt{61}+10}{4}$$

Divide $10+2\sqrt{61}$ by $4$.

$$x=\frac{\sqrt{61}+5}{2}$$

Now solve the equation $x=\frac{10±2\sqrt{61}}{4}$ when $±$ is minus. Subtract $2\sqrt{61}$ from $10$.

$$x=\frac{10-2\sqrt{61}}{4}$$

Divide $10-2\sqrt{61}$ by $4$.

$$x=\frac{5-\sqrt{61}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{61}+5}{2}$$ $$x=\frac{5-\sqrt{61}}{2}$$

Steps for Completing the Square

Subtract $x$ from both sides.

$$2x^{2}-9x-18-x=0$$

Combine $-9x$ and $-x$ to get $-10x$.

$$2x^{2}-10x-18=0$$

Add $18$ to both sides. Anything plus zero gives itself.

$$2x^{2}-10x=18$$

Divide both sides by $2$.

$$\frac{2x^{2}-10x}{2}=\frac{18}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}+\left(-\frac{10}{2}\right)x=\frac{18}{2}$$

Divide $-10$ by $2$.

$$x^{2}-5x=\frac{18}{2}$$

Divide $18$ by $2$.

$$x^{2}-5x=9$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=9+\left(-\frac{5}{2}\right)^{2}$$

Square $-\frac{5}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=9+\frac{25}{4}$$

Add $9$ to $\frac{25}{4}$.

$$x^{2}-5x+\frac{25}{4}=\frac{61}{4}$$

Factor $x^{2}-5x+\frac{25}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{61}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{61}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{61}}{2}$$ $$x-\frac{5}{2}=-\frac{\sqrt{61}}{2}$$

Add $\frac{5}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{61}+5}{2}$$ $$x=\frac{5-\sqrt{61}}{2}$$