How do you find the zeros of # y = -2x^2 - 3x +12 # using the quadratic formula?

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Answer 1 · 2015-12-02T08:25:55

Set #a#, #b#, and #c# appropriately and apply the quadratic formula to obtain the zeros at #x = (-3+sqrt(105))/4# and #x = (-3-sqrt(105))/4#

The quadratic formula states that

#ax^2 + bx + c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)#

In this case, we have
#a = -2#
#b = -3#
#c = 12#

Thus the zeros, that is, the solutions to #-2x^2 - 3x + 12 = 0# are

#x = (-(-3)+-sqrt((-3)^2-4(-2)(12)))/(2(-2))#

#=(3 +-sqrt(9 + 96))/(-4)#

#= -(3+-sqrt(105))/4#

#= (-3+-sqrt(105))/4#

So
#x = (-3+sqrt(105))/4#
or
#x = (-3-sqrt(105))/4#