Solve () (x + 2)(x - 5) = 4x | 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

$$( x + 2 ) ( x - 5 ) = 4 x$$

Solve for x

$x = \frac{\sqrt{89} + 7}{2} \approx 8.216990566$
$x=\frac{7-\sqrt{89}}{2}\approx -1.216990566$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $x+2$ by $x-5$ and combine like terms.

$$x^{2}-3x-10=4x$$

Subtract $4x$ from both sides.

$$x^{2}-3x-10-4x=0$$

Combine $-3x$ and $-4x$ to get $-7x$.

$$x^{2}-7x-10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-7$ for $b$, and $-10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-10\right)}}{2}$$

Square $-7$.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-10\right)}}{2}$$

Multiply $-4$ times $-10$.

$$x=\frac{-\left(-7\right)±\sqrt{49+40}}{2}$$

Add $49$ to $40$.

$$x=\frac{-\left(-7\right)±\sqrt{89}}{2}$$

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

$$x=\frac{7±\sqrt{89}}{2}$$

Now solve the equation $x=\frac{7±\sqrt{89}}{2}$ when $±$ is plus. Add $7$ to $\sqrt{89}$.

$$x=\frac{\sqrt{89}+7}{2}$$

Now solve the equation $x=\frac{7±\sqrt{89}}{2}$ when $±$ is minus. Subtract $\sqrt{89}$ from $7$.

$$x=\frac{7-\sqrt{89}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{89}+7}{2}$$ $$x=\frac{7-\sqrt{89}}{2}$$

Steps for Completing the Square

Use the distributive property to multiply $x+2$ by $x-5$ and combine like terms.

$$x^{2}-3x-10=4x$$

Subtract $4x$ from both sides.

$$x^{2}-3x-10-4x=0$$

Combine $-3x$ and $-4x$ to get $-7x$.

$$x^{2}-7x-10=0$$

Add $10$ to both sides. Anything plus zero gives itself.

$$x^{2}-7x=10$$

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.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=10+\left(-\frac{7}{2}\right)^{2}$$

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

$$x^{2}-7x+\frac{49}{4}=10+\frac{49}{4}$$

Add $10$ to $\frac{49}{4}$.

$$x^{2}-7x+\frac{49}{4}=\frac{89}{4}$$

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

$$\left(x-\frac{7}{2}\right)^{2}=\frac{89}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{89}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{89}}{2}$$ $$x-\frac{7}{2}=-\frac{\sqrt{89}}{2}$$

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

$$x=\frac{\sqrt{89}+7}{2}$$ $$x=\frac{7-\sqrt{89}}{2}$$