How do you solve #0= -2x^2 -8x +9# using the quadratic formula?

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Answer 1 · 2016-03-23T13:31:04

#x_1=-2-sqrt(34)/2#

#x_2=-2+sqrt(34)/2#

To simply the computation rewrite equation as:

#-2x^2-8x+9=0#

Moltiply both LHS and RHS by #-1# to obtain

#2x^2+8x-9=0#

#x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)#

with: #a=2#, #b=8# and #c=-9#

#x_(1,2)=(-8+-sqrt(64+72))/4=(-8+-sqrt(136))/4=#
#=-2+-sqrt(2^3*17)/4=-2+-cancel(2)sqrt(2*17)/cancel(4)^2=#
#=-2+-sqrt(34)/2#

#x_1=-2-sqrt(34)/2#

#x_2=-2+sqrt(34)/2#

graph{2x^2+8x-9 [-7.023, 7.024, -3.51, 3.513]}