Solve 5n^2-165n+1300=0

$$5 n ^ { 2 } - 165 n + 1300 = 0$$

Solve for n

$$n=13$$

$$n=20$$

Steps Using the Quadratic Formula

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.

$$5n^{2}-165n+1300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -165 for b, and 1300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-165\right)±\sqrt{\left(-165\right)^{2}-4\times 5\times 1300}}{2\times 5}$$

Square -165.

$$n=\frac{-\left(-165\right)±\sqrt{27225-4\times 5\times 1300}}{2\times 5}$$

Multiply -4 times 5.

$$n=\frac{-\left(-165\right)±\sqrt{27225-20\times 1300}}{2\times 5}$$

Multiply -20 times 1300.

$$n=\frac{-\left(-165\right)±\sqrt{27225-26000}}{2\times 5}$$

Add 27225 to -26000.

$$n=\frac{-\left(-165\right)±\sqrt{1225}}{2\times 5}$$

Take the square root of 1225.

$$n=\frac{-\left(-165\right)±35}{2\times 5}$$

The opposite of -165 is 165.

$$n=\frac{165±35}{2\times 5}$$

Multiply 2 times 5.

$$n=\frac{165±35}{10}$$

Now solve the equation n=\frac{165±35}{10} when ± is plus. Add 165 to 35.

$$n=\frac{200}{10}$$

Divide 200 by 10.

$$n=20$$

Now solve the equation n=\frac{165±35}{10} when ± is minus. Subtract 35 from 165.

$$n=\frac{130}{10}$$

Divide 130 by 10.

$$n=13$$

The equation is now solved.

$$n=20 n=13$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$5n^{2}-165n+1300=0$$

Subtract 1300 from both sides of the equation.

$$5n^{2}-165n+1300-1300=-1300$$

Subtracting 1300 from itself leaves 0.

$$5n^{2}-165n=-1300$$

Divide both sides by 5.

$$\frac{5n^{2}-165n}{5}=-\frac{1300}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$n^{2}+\left(-\frac{165}{5}\right)n=-\frac{1300}{5}$$

Divide -165 by 5.

$$n^{2}-33n=-\frac{1300}{5}$$

Divide -1300 by 5.

$$n^{2}-33n=-260$$

Divide -33, the coefficient of the x term, by 2 to get -\frac{33}{2}. Then add the square of -\frac{33}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-33n+\left(-\frac{33}{2}\right)^{2}=-260+\left(-\frac{33}{2}\right)^{2}$$

Square -\frac{33}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-33n+\frac{1089}{4}=-260+\frac{1089}{4}$$

Add -260 to \frac{1089}{4}.

$$n^{2}-33n+\frac{1089}{4}=\frac{49}{4}$$

Factor n^{2}-33n+\frac{1089}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{33}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{33}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$n-\frac{33}{2}=\frac{7}{2} n-\frac{33}{2}=-\frac{7}{2}$$

Add \frac{33}{2} to both sides of the equation.

$$n=20 n=13$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 5

$$x ^ 2 -33x +260 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 33 rs = 260$$

Two numbers r and s sum up to 33 exactly when the average of the two numbers is \frac{1}{2}*33 = \frac{33}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{33}{2} - u s = \frac{33}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 260

$$(\frac{33}{2} - u) (\frac{33}{2} + u) = 260$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{1089}{4} - u^2 = 260$$

Simplify the expression by subtracting \frac{1089}{4} on both sides

$$-u^2 = 260-\frac{1089}{4} = -\frac{49}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{33}{2} - \frac{7}{2} = 13 s = \frac{33}{2} + \frac{7}{2} = 20$$