Solve 2x^2+x-3-18=0

$$2 x ^ { 2 } + x - 3 - 18 = 0$$

Solve for x

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

$$x=3$$

Steps Using Factoring By Grouping

Subtract 18 from -3 to get -21.

$$2x^{2}+x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-21\right)=-42$$

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 -42.

$$-1,42 -2,21 -3,14 -6,7$$

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 1.

$$a=-6 b=7$$

Rewrite 2x^{2}+x-21 as \left(2x^{2}-6x\right)+\left(7x-21\right).

$$\left(2x^{2}-6x\right)+\left(7x-21\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(x-3\right)+7\left(x-3\right)$$

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

$$\left(x-3\right)\left(2x+7\right)$$

To find equation solutions, solve x-3=0 and 2x+7=0.

$$x=3 x=-\frac{7}{2}$$

Steps Using the Quadratic Formula

Subtract 18 from -3 to get -21.

$$2x^{2}+x-21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-21\right)}}{2\times 2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-21\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-1±\sqrt{1-8\left(-21\right)}}{2\times 2}$$

Multiply -8 times -21.

$$x=\frac{-1±\sqrt{1+168}}{2\times 2}$$

Add 1 to 168.

$$x=\frac{-1±\sqrt{169}}{2\times 2}$$

Take the square root of 169.

$$x=\frac{-1±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-1±13}{4}$$

Now solve the equation x=\frac{-1±13}{4} when ± is plus. Add -1 to 13.

$$x=\frac{12}{4}$$

Divide 12 by 4.

$$x=3$$

Now solve the equation x=\frac{-1±13}{4} when ± is minus. Subtract 13 from -1.

$$x=-\frac{14}{4}$$

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{7}{2}$$

The equation is now solved.

$$x=3 x=-\frac{7}{2}$$

Steps for Completing the Square

Subtract 18 from -3 to get -21.

$$2x^{2}+x-21=0$$

Add 21 to both sides. Anything plus zero gives itself.

$$2x^{2}+x=21$$

Divide both sides by 2.

$$\frac{2x^{2}+x}{2}=\frac{21}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{1}{2}x=\frac{21}{2}$$

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.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{21}{2}+\left(\frac{1}{4}\right)^{2}$$

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

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{21}{2}+\frac{1}{16}$$

Add \frac{21}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{169}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{13}{4} x+\frac{1}{4}=-\frac{13}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=3 x=-\frac{7}{2}$$