How do you solve #-10x^2 + 30x - 20 = 0.#?

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Answer 1 · 2018-05-27T04:35:11

#x= 2, 1#

given, #-10x^2 + 30x -20=0#
Multiplying both sides with #-1#
we get,
#10x^2 - 30x +20=0#

By Using quadratic formula #{x= (-b +- sqrt(b^2-4ac))/(2a)}#
we get,

#rArr x= (30 +- sqrt(30^2 - 4*10*20))/(2*10)#
#rArr x = (30 +- 10)/20#
#rArr x = 40/20 , x = 20/20#
Thus we get the value of #x# as #x = 2,1#

Answer 2 · 2018-05-27T04:39:43

#x=2#
#x=1#

Given -

#-10x^2+30x-20=0#

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

Dividing both sides by 10 we get

#x^2-3x+2=0#

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

#x(x-1)-2(x-1)=0#

#(x-2)(x-1)=0#

#x-2=0#

#x=2#

#x-1=0#

#x=1#