Solve 13n^2-41n-120=0

$$13 n ^ { 2 } - 41 n - 120 = 0$$

Solve for n

$$n = -\frac{24}{13} = -1\frac{11}{13} \approx -1.846153846$$

$$n=5$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 13n^{2}+an+bn-120. To find a and b, set up a system to be solved.

$$a+b=-41 ab=13\left(-120\right)=-1560$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -1560.

$$1,-1560 2,-780 3,-520 4,-390 5,-312 6,-260 8,-195 10,-156 12,-130 13,-120 15,-104 20,-78 24,-65 26,-60 30,-52 39,-40$$

Calculate the sum for each pair.

$$1-1560=-1559 2-780=-778 3-520=-517 4-390=-386 5-312=-307 6-260=-254 8-195=-187 10-156=-146 12-130=-118 13-120=-107 15-104=-89 20-78=-58 24-65=-41 26-60=-34 30-52=-22 39-40=-1$$

The solution is the pair that gives sum -41.

$$a=-65 b=24$$

Rewrite 13n^{2}-41n-120 as \left(13n^{2}-65n\right)+\left(24n-120\right).

$$\left(13n^{2}-65n\right)+\left(24n-120\right)$$

Factor out 13n in the first and 24 in the second group.

$$13n\left(n-5\right)+24\left(n-5\right)$$

Factor out common term n-5 by using distributive property.

$$\left(n-5\right)\left(13n+24\right)$$

To find equation solutions, solve n-5=0 and 13n+24=0.

$$n=5 n=-\frac{24}{13}$$

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.

$$13n^{2}-41n-120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 13 for a, -41 for b, and -120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 13\left(-120\right)}}{2\times 13}$$

Square -41.

$$n=\frac{-\left(-41\right)±\sqrt{1681-4\times 13\left(-120\right)}}{2\times 13}$$

Multiply -4 times 13.

$$n=\frac{-\left(-41\right)±\sqrt{1681-52\left(-120\right)}}{2\times 13}$$

Multiply -52 times -120.

$$n=\frac{-\left(-41\right)±\sqrt{1681+6240}}{2\times 13}$$

Add 1681 to 6240.

$$n=\frac{-\left(-41\right)±\sqrt{7921}}{2\times 13}$$

Take the square root of 7921.

$$n=\frac{-\left(-41\right)±89}{2\times 13}$$

The opposite of -41 is 41.

$$n=\frac{41±89}{2\times 13}$$

Multiply 2 times 13.

$$n=\frac{41±89}{26}$$

Now solve the equation n=\frac{41±89}{26} when ± is plus. Add 41 to 89.

$$n=\frac{130}{26}$$

Divide 130 by 26.

$$n=5$$

Now solve the equation n=\frac{41±89}{26} when ± is minus. Subtract 89 from 41.

$$n=-\frac{48}{26}$$

Reduce the fraction \frac{-48}{26} to lowest terms by extracting and canceling out 2.

$$n=-\frac{24}{13}$$

The equation is now solved.

$$n=5 n=-\frac{24}{13}$$

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.

$$13n^{2}-41n-120=0$$

Add 120 to both sides of the equation.

$$13n^{2}-41n-120-\left(-120\right)=-\left(-120\right)$$

Subtracting -120 from itself leaves 0.

$$13n^{2}-41n=-\left(-120\right)$$

Subtract -120 from 0.

$$13n^{2}-41n=120$$

Divide both sides by 13.

$$\frac{13n^{2}-41n}{13}=\frac{120}{13}$$

Dividing by 13 undoes the multiplication by 13.

$$n^{2}-\frac{41}{13}n=\frac{120}{13}$$

Divide -\frac{41}{13}, the coefficient of the x term, by 2 to get -\frac{41}{26}. Then add the square of -\frac{41}{26} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{41}{13}n+\left(-\frac{41}{26}\right)^{2}=\frac{120}{13}+\left(-\frac{41}{26}\right)^{2}$$

Square -\frac{41}{26} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{41}{13}n+\frac{1681}{676}=\frac{120}{13}+\frac{1681}{676}$$

Add \frac{120}{13} to \frac{1681}{676} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{41}{13}n+\frac{1681}{676}=\frac{7921}{676}$$

Factor n^{2}-\frac{41}{13}n+\frac{1681}{676}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{41}{26}\right)^{2}=\frac{7921}{676}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{41}{26}\right)^{2}}=\sqrt{\frac{7921}{676}}$$

Simplify.

$$n-\frac{41}{26}=\frac{89}{26} n-\frac{41}{26}=-\frac{89}{26}$$

Add \frac{41}{26} to both sides of the equation.

$$n=5 n=-\frac{24}{13}$$

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 13

$$x ^ 2 -\frac{41}{13}x -\frac{120}{13} = 0$$

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

$$r + s = \frac{41}{13} rs = -\frac{120}{13}$$

Two numbers r and s sum up to \frac{41}{13} exactly when the average of the two numbers is \frac{1}{2}*\frac{41}{13} = \frac{41}{26}. 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>

$$r = \frac{41}{26} - u s = \frac{41}{26} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{120}{13}

$$(\frac{41}{26} - u) (\frac{41}{26} + u) = -\frac{120}{13}$$

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

$$\frac{1681}{676} - u^2 = -\frac{120}{13}$$

Simplify the expression by subtracting \frac{1681}{676} on both sides

$$-u^2 = -\frac{120}{13}-\frac{1681}{676} = -\frac{7921}{676}$$

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

$$u^2 = \frac{7921}{676} u = \pm\sqrt{\frac{7921}{676}} = \pm \frac{89}{26}$$

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

$$r =\frac{41}{26} - \frac{89}{26} = -1.846 s = \frac{41}{26} + \frac{89}{26} = 5$$