Solve 18x^2-9x-5=0

$$18 { x }^{ 2 } -9x-5 = 0$$

Solve for x

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

$$x=\frac{5}{6}\approx 0.833333333$$

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 18x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-9 ab=18\left(-5\right)=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -9.

$$a=-15 b=6$$

Rewrite 18x^{2}-9x-5 as \left(18x^{2}-15x\right)+\left(6x-5\right).

$$\left(18x^{2}-15x\right)+\left(6x-5\right)$$

Factor out 3x in 18x^{2}-15x.

$$3x\left(6x-5\right)+6x-5$$

Factor out common term 6x-5 by using distributive property.

$$\left(6x-5\right)\left(3x+1\right)$$

To find equation solutions, solve 6x-5=0 and 3x+1=0.

$$x=\frac{5}{6} x=-\frac{1}{3}$$

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.

$$18x^{2}-9x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, -9 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 18\left(-5\right)}}{2\times 18}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 18\left(-5\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-\left(-9\right)±\sqrt{81-72\left(-5\right)}}{2\times 18}$$

Multiply -72 times -5.

$$x=\frac{-\left(-9\right)±\sqrt{81+360}}{2\times 18}$$

Add 81 to 360.

$$x=\frac{-\left(-9\right)±\sqrt{441}}{2\times 18}$$

Take the square root of 441.

$$x=\frac{-\left(-9\right)±21}{2\times 18}$$

The opposite of -9 is 9.

$$x=\frac{9±21}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{9±21}{36}$$

Now solve the equation x=\frac{9±21}{36} when ± is plus. Add 9 to 21.

$$x=\frac{30}{36}$$

Reduce the fraction \frac{30}{36} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{6}$$

Now solve the equation x=\frac{9±21}{36} when ± is minus. Subtract 21 from 9.

$$x=-\frac{12}{36}$$

Reduce the fraction \frac{-12}{36} to lowest terms by extracting and canceling out 12.

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

The equation is now solved.

$$x=\frac{5}{6} x=-\frac{1}{3}$$

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.

$$18x^{2}-9x-5=0$$

Add 5 to both sides of the equation.

$$18x^{2}-9x-5-\left(-5\right)=-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$18x^{2}-9x=-\left(-5\right)$$

Subtract -5 from 0.

$$18x^{2}-9x=5$$

Divide both sides by 18.

$$\frac{18x^{2}-9x}{18}=\frac{5}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\left(-\frac{9}{18}\right)x=\frac{5}{18}$$

Reduce the fraction \frac{-9}{18} to lowest terms by extracting and canceling out 9.

$$x^{2}-\frac{1}{2}x=\frac{5}{18}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{5}{18}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{5}{18}+\frac{1}{16}$$

Add \frac{5}{18} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{49}{144}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{49}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{144}}$$

Simplify.

$$x-\frac{1}{4}=\frac{7}{12} x-\frac{1}{4}=-\frac{7}{12}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{5}{6} x=-\frac{1}{3}$$