How do you solve by completing the square #2x^2-6x-20=0#?

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Answer 1 · 2018-03-23T18:28:11

#x_1=-2# and #x_2=5#

#2x^2-6x-20=0#

#2*(x^2-3x-10)=0#

#2*(x^2-3x+9/4-49/4)=0#

#2*((x-3/2)^2-(7/2)^2)=0#

#2(x-3/2+7/2)(x-3/2-7/2)=0#

#2(x+2)(x-5)=0#

Thus, #x_1=-2# and #x_2=5#