Solve 6x^2+5x-6=0

$$6 { x }^{ 2 } +5x-6=0$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=6\left(-6\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 5.

$$a=-4 b=9$$

Rewrite 6x^{2}+5x-6 as \left(6x^{2}-4x\right)+\left(9x-6\right).

$$\left(6x^{2}-4x\right)+\left(9x-6\right)$$

Factor out 2x in the first and 3 in the second group.

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

Factor out common term 3x-2 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

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.

$$6x^{2}+5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 6\left(-6\right)}}{2\times 6}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 6\left(-6\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-5±\sqrt{25-24\left(-6\right)}}{2\times 6}$$

Multiply -24 times -6.

$$x=\frac{-5±\sqrt{25+144}}{2\times 6}$$

Add 25 to 144.

$$x=\frac{-5±\sqrt{169}}{2\times 6}$$

Take the square root of 169.

$$x=\frac{-5±13}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-5±13}{12}$$

Now solve the equation x=\frac{-5±13}{12} when ± is plus. Add -5 to 13.

$$x=\frac{8}{12}$$

Reduce the fraction \frac{8}{12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-5±13}{12} when ± is minus. Subtract 13 from -5.

$$x=-\frac{18}{12}$$

Reduce the fraction \frac{-18}{12} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

Steps for Completing the Square

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.

$$6x^{2}+5x-6=0$$

Add 6 to both sides of the equation.

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

Subtracting -6 from itself leaves 0.

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

Subtract -6 from 0.

$$6x^{2}+5x=6$$

Divide both sides by 6.

$$\frac{6x^{2}+5x}{6}=\frac{6}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{5}{6}x=\frac{6}{6}$$

Divide 6 by 6.

$$x^{2}+\frac{5}{6}x=1$$

Divide \frac{5}{6}, the coefficient of the x term, by 2 to get \frac{5}{12}. Then add the square of \frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{6}x+\left(\frac{5}{12}\right)^{2}=1+\left(\frac{5}{12}\right)^{2}$$

Square \frac{5}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{6}x+\frac{25}{144}=1+\frac{25}{144}$$

Add 1 to \frac{25}{144}.

$$x^{2}+\frac{5}{6}x+\frac{25}{144}=\frac{169}{144}$$

Factor x^{2}+\frac{5}{6}x+\frac{25}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{12}\right)^{2}=\frac{169}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{12}\right)^{2}}=\sqrt{\frac{169}{144}}$$

Simplify.

$$x+\frac{5}{12}=\frac{13}{12} x+\frac{5}{12}=-\frac{13}{12}$$

Subtract \frac{5}{12} from both sides of the equation.

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