Solve 2*(a^2-9)+a=15

$$2 \cdot ( a ^ { 2 } - 9 ) + a = 15$$

Solve for a

$$a = \frac{\sqrt{265} - 1}{4} \approx 3.819705149$$

$$a=\frac{-\sqrt{265}-1}{4}\approx -4.319705149$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by a^{2}-9.

$$2a^{2}-18+a=15$$

Subtract 15 from both sides.

$$2a^{2}-18+a-15=0$$

Subtract 15 from -18 to get -33.

$$2a^{2}-33+a=0$$

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.

$$2a^{2}+a-33=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -33 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 1.

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

Multiply -4 times 2.

$$a=\frac{-1±\sqrt{1-8\left(-33\right)}}{2\times 2}$$

Multiply -8 times -33.

$$a=\frac{-1±\sqrt{1+264}}{2\times 2}$$

Add 1 to 264.

$$a=\frac{-1±\sqrt{265}}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{-1±\sqrt{265}}{4}$$

Now solve the equation a=\frac{-1±\sqrt{265}}{4} when ± is plus. Add -1 to \sqrt{265}.

$$a=\frac{\sqrt{265}-1}{4}$$

Now solve the equation a=\frac{-1±\sqrt{265}}{4} when ± is minus. Subtract \sqrt{265} from -1.

$$a=\frac{-\sqrt{265}-1}{4}$$

The equation is now solved.

$$a=\frac{\sqrt{265}-1}{4} a=\frac{-\sqrt{265}-1}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 2 by a^{2}-9.

$$2a^{2}-18+a=15$$

Add 18 to both sides.

$$2a^{2}+a=15+18$$

Add 15 and 18 to get 33.

$$2a^{2}+a=33$$

Divide both sides by 2.

$$\frac{2a^{2}+a}{2}=\frac{33}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

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.

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

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

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

Add \frac{33}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{1}{2}a+\frac{1}{16}=\frac{265}{16}$$

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}.

$$\left(a+\frac{1}{4}\right)^{2}=\frac{265}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{1}{4}\right)^{2}}=\sqrt{\frac{265}{16}}$$

Simplify.

$$a+\frac{1}{4}=\frac{\sqrt{265}}{4} a+\frac{1}{4}=-\frac{\sqrt{265}}{4}$$

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

$$a=\frac{\sqrt{265}-1}{4} a=\frac{-\sqrt{265}-1}{4}$$