Solve (3x-10)(x+1)=10

$$(3x-10)(x+1)=10$$

Solve for x

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

$$x=4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x-10 by x+1 and combine like terms.

$$3x^{2}-7x-10=10$$

Subtract 10 from both sides.

$$3x^{2}-7x-10-10=0$$

Subtract 10 from -10 to get -20.

$$3x^{2}-7x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -7 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\left(-20\right)}}{2\times 3}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 3\left(-20\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-12\left(-20\right)}}{2\times 3}$$

Multiply -12 times -20.

$$x=\frac{-\left(-7\right)±\sqrt{49+240}}{2\times 3}$$

Add 49 to 240.

$$x=\frac{-\left(-7\right)±\sqrt{289}}{2\times 3}$$

Take the square root of 289.

$$x=\frac{-\left(-7\right)±17}{2\times 3}$$

The opposite of -7 is 7.

$$x=\frac{7±17}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{7±17}{6}$$

Now solve the equation x=\frac{7±17}{6} when ± is plus. Add 7 to 17.

$$x=\frac{24}{6}$$

Divide 24 by 6.

$$x=4$$

Now solve the equation x=\frac{7±17}{6} when ± is minus. Subtract 17 from 7.

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

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

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

The equation is now solved.

$$x=4 x=-\frac{5}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 3x-10 by x+1 and combine like terms.

$$3x^{2}-7x-10=10$$

Add 10 to both sides.

$$3x^{2}-7x=10+10$$

Add 10 and 10 to get 20.

$$3x^{2}-7x=20$$

Divide both sides by 3.

$$\frac{3x^{2}-7x}{3}=\frac{20}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{7}{3}x=\frac{20}{3}$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=\frac{20}{3}+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{20}{3}+\frac{49}{36}$$

Add \frac{20}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{289}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{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}.

$$\left(x-\frac{7}{6}\right)^{2}=\frac{289}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{\frac{289}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{17}{6} x-\frac{7}{6}=-\frac{17}{6}$$

Add \frac{7}{6} to both sides of the equation.

$$x=4 x=-\frac{5}{3}$$