How to solve (7-3x)(x+2)=4

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 (7-3x)(x+2)=4 | Equatix

Question

$$(7-3x)(x+2)=4$$

Solve for x

$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$
$x=2$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $7-3x$ by $x+2$ and combine like terms.

$$x+14-3x^{2}=4$$

Subtract $4$ from both sides.

$$x+14-3x^{2}-4=0$$

Subtract $4$ from $14$ to get $10$.

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

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.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-3$ for $a$, $1$ for $b$, and $10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-3\right)\times 10}}{2\left(-3\right)}$$

Square $1$.

$$x=\frac{-1±\sqrt{1-4\left(-3\right)\times 10}}{2\left(-3\right)}$$

Multiply $-4$ times $-3$.

$$x=\frac{-1±\sqrt{1+12\times 10}}{2\left(-3\right)}$$

Multiply $12$ times $10$.

$$x=\frac{-1±\sqrt{1+120}}{2\left(-3\right)}$$

Add $1$ to $120$.

$$x=\frac{-1±\sqrt{121}}{2\left(-3\right)}$$

Take the square root of $121$.

$$x=\frac{-1±11}{2\left(-3\right)}$$

Multiply $2$ times $-3$.

$$x=\frac{-1±11}{-6}$$

Now solve the equation $x=\frac{-1±11}{-6}$ when $±$ is plus. Add $-1$ to $11$.

$$x=\frac{10}{-6}$$

Reduce the fraction $\frac{10}{-6}$ to lowest terms by extracting and canceling out $2$.

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

Now solve the equation $x=\frac{-1±11}{-6}$ when $±$ is minus. Subtract $11$ from $-1$.

$$x=-\frac{12}{-6}$$

Divide $-12$ by $-6$.

$$x=2$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply $7-3x$ by $x+2$ and combine like terms.

$$x+14-3x^{2}=4$$

Subtract $14$ from both sides.

$$x-3x^{2}=4-14$$

Subtract $14$ from $4$ to get $-10$.

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

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

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

Divide both sides by $-3$.

$$\frac{-3x^{2}+x}{-3}=-\frac{10}{-3}$$

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

$$x^{2}+\frac{1}{-3}x=-\frac{10}{-3}$$

Divide $1$ by $-3$.

$$x^{2}-\frac{1}{3}x=-\frac{10}{-3}$$

Divide $-10$ by $-3$.

$$x^{2}-\frac{1}{3}x=\frac{10}{3}$$

Divide $-\frac{1}{3}$, the coefficient of the $x$ term, by $2$ to get $-\frac{1}{6}$. Then add the square of $-\frac{1}{6}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{10}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square $-\frac{1}{6}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{10}{3}+\frac{1}{36}$$

Add $\frac{10}{3}$ to $\frac{1}{36}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{121}{36}$$

Factor $x^{2}-\frac{1}{3}x+\frac{1}{36}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{121}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{11}{6}$$ $$x-\frac{1}{6}=-\frac{11}{6}$$

Add $\frac{1}{6}$ to both sides of the equation.

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

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Example

Question

Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square