Solve 6\timesx+x\times(x-13)=18 | 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

$$6 \times x+x \times (x-13)=18$$

Solve for x

$x=-2$
$x=9$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $x$ by $x-13$.

$$6x+x^{2}-13x=18$$

Combine $6x$ and $-13x$ to get $-7x$.

$$-7x+x^{2}=18$$

Subtract $18$ from both sides.

$$-7x+x^{2}-18=0$$

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.

$$x^{2}-7x-18=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-7$ for $b$, and $-18$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-18\right)}}{2}$$

Square $-7$.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-18\right)}}{2}$$

Multiply $-4$ times $-18$.

$$x=\frac{-\left(-7\right)±\sqrt{49+72}}{2}$$

Add $49$ to $72$.

$$x=\frac{-\left(-7\right)±\sqrt{121}}{2}$$

Take the square root of $121$.

$$x=\frac{-\left(-7\right)±11}{2}$$

The opposite of $-7$ is $7$.

$$x=\frac{7±11}{2}$$

Now solve the equation $x=\frac{7±11}{2}$ when $±$ is plus. Add $7$ to $11$.

$$x=\frac{18}{2}$$

Divide $18$ by $2$.

$$x=9$$

Now solve the equation $x=\frac{7±11}{2}$ when $±$ is minus. Subtract $11$ from $7$.

$$x=-\frac{4}{2}$$

Divide $-4$ by $2$.

$$x=-2$$

The equation is now solved.

$$x=9$$ $$x=-2$$

Steps for Completing the Square

Use the distributive property to multiply $x$ by $x-13$.

$$6x+x^{2}-13x=18$$

Combine $6x$ and $-13x$ to get $-7x$.

$$-7x+x^{2}=18$$

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 $x^{2}+bx=c$.

$$x^{2}-7x=18$$

Divide $-7$, the coefficient of the $x$ term, by $2$ to get $-\frac{7}{2}$. Then add the square of $-\frac{7}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=18+\left(-\frac{7}{2}\right)^{2}$$

Square $-\frac{7}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=18+\frac{49}{4}$$

Add $18$ to $\frac{49}{4}$.

$$x^{2}-7x+\frac{49}{4}=\frac{121}{4}$$

Factor $x^{2}-7x+\frac{49}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{11}{2}$$ $$x-\frac{7}{2}=-\frac{11}{2}$$

Add $\frac{7}{2}$ to both sides of the equation.

$$x=9$$ $$x=-2$$