Solve 2a^2=a+4

$$2 a ^ { 2 } = a + 4$$

Solve for a

$$a = \frac{\sqrt{33} + 1}{4} \approx 1.686140662$$

$$a=\frac{1-\sqrt{33}}{4}\approx -1.186140662$$

Steps Using the Quadratic Formula

Subtract a from both sides.

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

Subtract 4 from both sides.

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

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

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

Multiply -4 times 2.

$$a=\frac{-\left(-1\right)±\sqrt{1-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$a=\frac{-\left(-1\right)±\sqrt{1+32}}{2\times 2}$$

Add 1 to 32.

$$a=\frac{-\left(-1\right)±\sqrt{33}}{2\times 2}$$

The opposite of -1 is 1.

$$a=\frac{1±\sqrt{33}}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{1±\sqrt{33}}{4}$$

Now solve the equation a=\frac{1±\sqrt{33}}{4} when ± is plus. Add 1 to \sqrt{33}.

$$a=\frac{\sqrt{33}+1}{4}$$

Now solve the equation a=\frac{1±\sqrt{33}}{4} when ± is minus. Subtract \sqrt{33} from 1.

$$a=\frac{1-\sqrt{33}}{4}$$

The equation is now solved.

$$a=\frac{\sqrt{33}+1}{4} a=\frac{1-\sqrt{33}}{4}$$

Steps for Completing the Square

Subtract a from both sides.

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

Divide both sides by 2.

$$\frac{2a^{2}-a}{2}=\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$a^{2}-\frac{1}{2}a=\frac{4}{2}$$

Divide 4 by 2.

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

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.

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

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

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

Add 2 to \frac{1}{16}.

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=\frac{33}{16}$$

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}.

$$\left(a-\frac{1}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{1}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$a-\frac{1}{4}=\frac{\sqrt{33}}{4} a-\frac{1}{4}=-\frac{\sqrt{33}}{4}$$

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

$$a=\frac{\sqrt{33}+1}{4} a=\frac{1-\sqrt{33}}{4}$$