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

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Answer 1 · 2015-08-08T20:34:58

#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#