How do you solve #14n^2 + 24n - 5004 = 0#?

Question

Kennedy Adams · Accepted Answer

Quadratic formula

The straightforward way is to use the quadratic formula: #n=(-b+-sqrt(b^2-4ac))/(2a)# This is for expressions of the form #an^2 + bn + c#, so in your case a=14, b=24, c=-5004 . A slightly more complicated way is to reverse foil, trying to factor out your a, b, and c so you can write you equation into #(xn + y)(sn +t) = 0#, where x, y, s, and t are all numbers. Then you can solve one #()# at a time. But if you have a calculator, the quadratic formula is the best way to go. It'll work every time.

Isaiah Anthony · Answer

#n = -6/7+-15/7sqrt(78)#

After completing the square, use the identity difference between the squares to find:

#0 = 7/2(14n^2+24n-5004)#

#color(white)(0) = 49n^2+84n-17514#

#color(white)(0) = (7n)^2+2(7n)(6)+36-17550#

#color(white)(0) = (7n)^2+2(7n)(6)+6^2-(15^2 * 78)#

#color(white)(0) = (7n+6)^2-(15sqrt(78))^2#

#color(white)(0) = ((7n+6)-15sqrt(78))((7n+6)+15sqrt(78))#

#color(white)(0) = (7n+6-15sqrt(78))(7n+6+15sqrt(78))#

Hence:

#7n = -6+-15sqrt(78)#

So:

#n = -6/7+-15/7sqrt(78)#

Nathan Axel · Answer

undefined To solve the quadratic equation 14n^2 + 24n - 5004 = 0, you can use the quadratic formula:

n = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 14, b = 24, and c = -5004.

Plugging these values into the formula:

n = (-24 ± √(24^2 - 4 * 14 * (-5004))) / (2 * 14)

n = (-24 ± √(576 + 280896)) / 28

n = (-24 ± √281472) / 28

n = (-24 ± 530.378) / 28

This gives two possible solutions:

n₁ = (-24 + 530.378) / 28 ≈ 18.372

n₂ = (-24 - 530.378) / 28 ≈ -18.778

So, the solutions to the equation are approximately 18.372 and -18.778.