Solve for n 4^(2n^2+2n)=8

$4^{2 n^{2} + 2 n} = 8$

Step 1

Create equivalent expressions in the equation that all have equal bases.

$2^{2 (2 n^{2} + 2 n)} = 2^{3}$

Step 2

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$2 (2 n^{2} + 2 n) = 3$

Step 3

Solve for $n$ .

Divide each term in $2 (2 n^{2} + 2 n) = 3$ by $2$ and simplify.

Divide each term in $2 (2 n^{2} + 2 n) = 3$ by $2$ .

$\frac{2 (2 n^{2} + 2 n)}{2} = \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (2 n^{2} + 2 n)}{2} = \frac{3}{2}$

Divide $2 n^{2} + 2 n$ by $1$ .

$2 n^{2} + 2 n = \frac{3}{2}$

Subtract $\frac{3}{2}$ from both sides of the equation.

$2 n^{2} + 2 n - \frac{3}{2} = 0$

Multiply through by the least common denominator $2$ , then simplify.

Apply the distributive property.

$2 (2 n^{2}) + 2 (2 n) + 2 (- \frac{3}{2}) = 0$

Simplify.

Multiply $2$ by $2$ .

$4 n^{2} + 2 (2 n) + 2 (- \frac{3}{2}) = 0$

Multiply $2$ by $2$ .

$4 n^{2} + 4 n + 2 (- \frac{3}{2}) = 0$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$4 n^{2} + 4 n + 2 (\frac{- 3}{2}) = 0$

Cancel the common factor.

$4 n^{2} + 4 n + 2 (\frac{- 3}{2}) = 0$

Rewrite the expression.

$4 n^{2} + 4 n - 3 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 4$ , $b = 4$ , and $c = - 3$ into the quadratic formula and solve for $n$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (4 \cdot - 3)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$n = \frac{- 4 \pm \sqrt{16 - 4 \cdot 4 \cdot - 3}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 3$ .

Multiply $- 4$ by $4$ .

$n = \frac{- 4 \pm \sqrt{16 - 16 \cdot - 3}}{2 \cdot 4}$

Multiply $- 16$ by $- 3$ .

$n = \frac{- 4 \pm \sqrt{16 + 48}}{2 \cdot 4}$

Add $16$ and $48$ .

$n = \frac{- 4 \pm \sqrt{64}}{2 \cdot 4}$

Rewrite $64$ as $8^{2}$ .

$n = \frac{- 4 \pm \sqrt{8^{2}}}{2 \cdot 4}$

Pull terms out from under the radical, assuming positive real numbers.

$n = \frac{- 4 \pm 8}{2 \cdot 4}$

Multiply $2$ by $4$ .

$n = \frac{- 4 \pm 8}{8}$

Simplify $\frac{- 4 \pm 8}{8}$ .

$n = \frac{- 1 \pm 2}{2}$

The final answer is the combination of both solutions.

$n = \frac{1}{2}, - \frac{3}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$n = \frac{1}{2}, - \frac{3}{2}$

Decimal Form:

$n = 0.5, - 1.5$

Mixed Number Form:

$n = \frac{1}{2}, - 1 \frac{1}{2}$

Solve for n 4^(2n^2+2n)=8

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