How do you solve #5x^2 - 10x - 12 = 0 #?

Question

Avery Alexander · Accepted Answer

Use the quadratic formula to find:

#x = 1 +- sqrt(85)/5# #5x^2-10x-12# is of the form #ax^2+bx+c# with #a=5#, #b=-10# and #c=-12# This has discriminant #Delta# given by the formula: #Delta = b^2-4ac = (-10)^2-(4xx5xx-12)# #= 100+240 = 340 = 2^2*85# This is positive, but not a perfect square, so the quadratic equation has a pair of irrational roots, given by the quadratic formula: #x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a)# #=(10+-sqrt(340))/10# #=(10+-2sqrt(85))/10# #=1+-sqrt(85)/5#

Savannah Anderson · Answer

undefined You can solve the quadratic equation 5x^2 - 10x - 12 = 0 using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 5, b = -10, and c = -12. Substitute these values into the formula:

x = (-(-10) ± √((-10)^2 - 4(5)(-12))) / (2(5))

x = (10 ± √(100 + 240)) / 10

x = (10 ± √340) / 10

x = (10 ± √(4 * 85)) / 10

x = (10 ± 2√85) / 10

Now, simplify the expression:

x = (5 ± √85) / 5

So, the solutions to the quadratic equation 5x^2 - 10x - 12 = 0 are:

x = (5 + √85) / 5
x = (5 - √85) / 5