Solve 3x^2+7x-10/x=0

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

Solve for x

$$x = -\frac{10}{3} = -3\frac{1}{3} \approx -3.333333333$$

$$x=1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=7 ab=3\left(-10\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=-3 b=10$$

Rewrite 3x^{2}+7x-10 as \left(3x^{2}-3x\right)+\left(10x-10\right).

$$\left(3x^{2}-3x\right)+\left(10x-10\right)$$

Factor out 3x in the first and 10 in the second group.

$$3x\left(x-1\right)+10\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(3x+10\right)$$

To find equation solutions, solve x-1=0 and 3x+10=0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3\left(-10\right)}}{2\times 3}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 3\left(-10\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-7±\sqrt{49-12\left(-10\right)}}{2\times 3}$$

Multiply -12 times -10.

$$x=\frac{-7±\sqrt{49+120}}{2\times 3}$$

Add 49 to 120.

$$x=\frac{-7±\sqrt{169}}{2\times 3}$$

Take the square root of 169.

$$x=\frac{-7±13}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-7±13}{6}$$

Now solve the equation x=\frac{-7±13}{6} when ± is plus. Add -7 to 13.

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

Divide 6 by 6.

$$x=1$$

Now solve the equation x=\frac{-7±13}{6} when ± is minus. Subtract 13 from -7.

$$x=-\frac{20}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{10}{3}$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

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

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

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

Divide both sides by 3.

$$\frac{3x^{2}+7x}{3}=\frac{10}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{7}{3}x=\frac{10}{3}$$

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.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=\frac{10}{3}+\left(\frac{7}{6}\right)^{2}$$

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

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{10}{3}+\frac{49}{36}$$

Add \frac{10}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{169}{36}$$

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}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{13}{6} x+\frac{7}{6}=-\frac{13}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

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