Solve Using the Quadratic Formula (5n+1)^2+2(5n+1)-3=0

${(5 n + 1)}^{2} + 2 (5 n + 1) - 3 = 0$

Step 1

Simplify the left side.

Simplify ${(5 n + 1)}^{2} + 2 (5 n + 1) - 3$ .

Simplify each term.

Apply the distributive property.

${(5 n + 1)}^{2} + 2 (5 n) + 2 \cdot 1 - 3 = 0$

Multiply $5$ by $2$ .

${(5 n + 1)}^{2} + 10 n + 2 \cdot 1 - 3 = 0$

Multiply $2$ by $1$ .

${(5 n + 1)}^{2} + 10 n + 2 - 3 = 0$

Subtract $3$ from $2$ .

${(5 n + 1)}^{2} + 10 n - 1 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 25$ , $b = 20$ , and $c = 0$ into the quadratic formula and solve for $n$ .

$\frac{- 20 \pm \sqrt{20^{2} - 4 \cdot (25 \cdot 0)}}{2 \cdot 25}$

Step 4

Simplify.

Simplify the numerator.

Raise $20$ to the power of $2$ .

$n = \frac{- 20 \pm \sqrt{400 - 4 \cdot 25 \cdot 0}}{2 \cdot 25}$

Multiply $- 4 \cdot 25 \cdot 0$ .

Multiply $- 4$ by $25$ .

$n = \frac{- 20 \pm \sqrt{400 - 100 \cdot 0}}{2 \cdot 25}$

Multiply $- 100$ by $0$ .

$n = \frac{- 20 \pm \sqrt{400 + 0}}{2 \cdot 25}$

Add $400$ and $0$ .

$n = \frac{- 20 \pm \sqrt{400}}{2 \cdot 25}$

Rewrite $400$ as $20^{2}$ .

$n = \frac{- 20 \pm \sqrt{20^{2}}}{2 \cdot 25}$

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

$n = \frac{- 20 \pm 20}{2 \cdot 25}$

Multiply $2$ by $25$ .

$n = \frac{- 20 \pm 20}{50}$

Simplify $\frac{- 20 \pm 20}{50}$ .

$n = \frac{- 2 \pm 2}{5}$

Step 5

The final answer is the combination of both solutions.

$n = 0, - \frac{4}{5}$