Solve 0=3x^2+2x-5

$$0 = 3 x ^ { 2 } + 2 x - 5$$

Solve for x

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

$$x=1$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}+2x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=2 ab=3\left(-5\right)=-15$$

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 -15.

$$-1,15 -3,5$$

Calculate the sum for each pair.

$$-1+15=14 -3+5=2$$

The solution is the pair that gives sum 2.

$$a=-3 b=5$$

Rewrite 3x^{2}+2x-5 as \left(3x^{2}-3x\right)+\left(5x-5\right).

$$\left(3x^{2}-3x\right)+\left(5x-5\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(x-1\right)+5\left(x-1\right)$$

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

$$\left(x-1\right)\left(3x+5\right)$$

To find equation solutions, solve x-1=0 and 3x+5=0.

$$x=1 x=-\frac{5}{3}$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}+2x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 2 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-2±\sqrt{4-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$x=\frac{-2±\sqrt{4+60}}{2\times 3}$$

Add 4 to 60.

$$x=\frac{-2±\sqrt{64}}{2\times 3}$$

Take the square root of 64.

$$x=\frac{-2±8}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-2±8}{6}$$

Now solve the equation x=\frac{-2±8}{6} when ± is plus. Add -2 to 8.

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

Divide 6 by 6.

$$x=1$$

Now solve the equation x=\frac{-2±8}{6} when ± is minus. Subtract 8 from -2.

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

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

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

The equation is now solved.

$$x=1 x=-\frac{5}{3}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}+2x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$3x^{2}+2x=5$$

Divide both sides by 3.

$$\frac{3x^{2}+2x}{3}=\frac{5}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{2}{3}x=\frac{5}{3}$$

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

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{5}{3}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{5}{3}+\frac{1}{9}$$

Add \frac{5}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{16}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{4}{3} x+\frac{1}{3}=-\frac{4}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=1 x=-\frac{5}{3}$$