How do you solve #2x ^ { 2} - 4x + 18= 0#?

1 Answer
Paramecium
Oct 30, 2017

No solution.

Explanation:

Divide both sides of the equation by 2.
#x^2-2x+9=0#
In this case, you cannot factor, so use the quadratic formula.
#x=(-b±sqrt(b^2-4ac))/2#
#x=(2±sqrt(4-4*1*9))/2#
#x=(2±sqrt(-32))/2#

However, #sqrt(-32)# does not exist, so no values of x satisfy the equation.

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