Solve -2=3x^2+5x-4

$$-2=3 { x }^{ 2 } +5x-4$$

Solve for x

$$x=-2$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

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

$$3x^{2}+5x-4=-2$$

Add 2 to both sides.

$$3x^{2}+5x-4+2=0$$

Add -4 and 2 to get -2.

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

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

$$a+b=5 ab=3\left(-2\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

Rewrite 3x^{2}+5x-2 as \left(3x^{2}-x\right)+\left(6x-2\right).

$$\left(3x^{2}-x\right)+\left(6x-2\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(3x-1\right)+2\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(x+2\right)$$

To find equation solutions, solve 3x-1=0 and x+2=0.

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

Steps Using the Quadratic Formula

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

$$3x^{2}+5x-4=-2$$

Add 2 to both sides.

$$3x^{2}+5x-4+2=0$$

Add -4 and 2 to get -2.

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

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

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

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 3\left(-2\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-5±\sqrt{25-12\left(-2\right)}}{2\times 3}$$

Multiply -12 times -2.

$$x=\frac{-5±\sqrt{25+24}}{2\times 3}$$

Add 25 to 24.

$$x=\frac{-5±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$x=\frac{-5±7}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-5±7}{6}$$

Now solve the equation x=\frac{-5±7}{6} when ± is plus. Add -5 to 7.

$$x=\frac{2}{6}$$

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

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

Now solve the equation x=\frac{-5±7}{6} when ± is minus. Subtract 7 from -5.

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

Divide -12 by 6.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

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

$$3x^{2}+5x-4=-2$$

Add 4 to both sides.

$$3x^{2}+5x=-2+4$$

Add -2 and 4 to get 2.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=\frac{2}{3}+\left(\frac{5}{6}\right)^{2}$$

Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{2}{3}+\frac{25}{36}$$

Add \frac{2}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{49}{36}$$

Factor x^{2}+\frac{5}{3}x+\frac{25}{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}.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x+\frac{5}{6}=\frac{7}{6} x+\frac{5}{6}=-\frac{7}{6}$$

Subtract \frac{5}{6} from both sides of the equation.

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