Solve 2a^2-4-a-11=0

$$2 a ^ { 2 } - 4 - a - 11 = 0$$

Solve for a

$$a = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$a=3$$

Steps Using Factoring By Grouping

Subtract 11 from -4 to get -15.

$$2a^{2}-15-a=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$2a^{2}-a-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2a^{2}+aa+ba-15. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-15\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

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

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

Factor out 2a in the first and 5 in the second group.

$$2a\left(a-3\right)+5\left(a-3\right)$$

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

$$\left(a-3\right)\left(2a+5\right)$$

To find equation solutions, solve a-3=0 and 2a+5=0.

$$a=3 a=-\frac{5}{2}$$

Steps Using the Quadratic Formula

Subtract 11 from -4 to get -15.

$$2a^{2}-15-a=0$$

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.

$$2a^{2}-a-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-15\right)}}{2\times 2}$$

Multiply -4 times 2.

$$a=\frac{-\left(-1\right)±\sqrt{1-8\left(-15\right)}}{2\times 2}$$

Multiply -8 times -15.

$$a=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 2}$$

Add 1 to 120.

$$a=\frac{-\left(-1\right)±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$a=\frac{-\left(-1\right)±11}{2\times 2}$$

The opposite of -1 is 1.

$$a=\frac{1±11}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{1±11}{4}$$

Now solve the equation a=\frac{1±11}{4} when ± is plus. Add 1 to 11.

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

Divide 12 by 4.

$$a=3$$

Now solve the equation a=\frac{1±11}{4} when ± is minus. Subtract 11 from 1.

$$a=-\frac{10}{4}$$

Reduce the fraction \frac{-10}{4} to lowest terms by extracting and canceling out 2.

$$a=-\frac{5}{2}$$

The equation is now solved.

$$a=3 a=-\frac{5}{2}$$

Steps for Completing the Square

Subtract 11 from -4 to get -15.

$$2a^{2}-15-a=0$$

Add 15 to both sides. Anything plus zero gives itself.

$$2a^{2}-a=15$$

Divide both sides by 2.

$$\frac{2a^{2}-a}{2}=\frac{15}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$a^{2}-\frac{1}{2}a=\frac{15}{2}$$

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.

$$a^{2}-\frac{1}{2}a+\left(-\frac{1}{4}\right)^{2}=\frac{15}{2}+\left(-\frac{1}{4}\right)^{2}$$

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

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=\frac{15}{2}+\frac{1}{16}$$

Add \frac{15}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=\frac{121}{16}$$

Factor a^{2}-\frac{1}{2}a+\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}.

$$\left(a-\frac{1}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$a-\frac{1}{4}=\frac{11}{4} a-\frac{1}{4}=-\frac{11}{4}$$

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

$$a=3 a=-\frac{5}{2}$$