Solve Using the Quadratic Formula 10n^2-35=65n

$10 n^{2} - 35 = 65 n$

Step 1

Subtract $65 n$ from both sides of the equation.

$10 n^{2} - 35 - 65 n = 0$

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 $a = 10$ , $b = - 65$ , and $c = - 35$ into the quadratic formula and solve for $n$ .

$\frac{65 \pm \sqrt{{(- 65)}^{2} - 4 \cdot (10 \cdot - 35)}}{2 \cdot 10}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 65$ to the power of $2$ .

$n = \frac{65 \pm \sqrt{4225 - 4 \cdot 10 \cdot - 35}}{2 \cdot 10}$

Multiply $- 4 \cdot 10 \cdot - 35$ .

Multiply $- 4$ by $10$ .

$n = \frac{65 \pm \sqrt{4225 - 40 \cdot - 35}}{2 \cdot 10}$

Multiply $- 40$ by $- 35$ .

$n = \frac{65 \pm \sqrt{4225 + 1400}}{2 \cdot 10}$

Add $4225$ and $1400$ .

$n = \frac{65 \pm \sqrt{5625}}{2 \cdot 10}$

Rewrite $5625$ as $75^{2}$ .

$n = \frac{65 \pm \sqrt{75^{2}}}{2 \cdot 10}$

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

$n = \frac{65 \pm 75}{2 \cdot 10}$

Multiply $2$ by $10$ .

$n = \frac{65 \pm 75}{20}$

Simplify $\frac{65 \pm 75}{20}$ .

$n = \frac{13 \pm 15}{4}$

Step 5

The final answer is the combination of both solutions.

$n = 7, - \frac{1}{2}$