How do you solve #2r^2-10r+12=0#?

1 Answer
Alan P.
Mar 27, 2015

#2r^2 - 10r + 12 = 0#
is of the general form for quadratic equations:
#ax^2 + bx + c = 0# with (#r# in place of #x#)

The general solution for values of #x# that satisfy this equation is given by the formula:
#(-b +- sqrt(b^2 - 4ac))/(2a)#

Substituting coefficients from the given quadratic we get:

#r_0 = (10 +- sqrt(100 -96))/4#
#= (10 +- 2)/4#

#rarr r_0 = 3# or #r_0 = 2#

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