How do you solve #2x ^ { 2} - 20= 3#?

1 Answer
smendyka
Apr 6, 2017

See the entire solution process below:

Explanation:

First, add #color(red)(20)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#2x^2 - 20 + color(red)(20) = 3 + color(red)(20)#

#2x^2 - 0 = 23#

#2x^2 = 23#

Now, divide each side of the equation by #color(red)(2)# to isolate the #x^2# term while keeping the equation balanced:

#(2x^2)/color(red)(2) = 23/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x^2)/cancel(color(red)(2)) = 23/2#

#x^2 = 23/2#

Now, take the square root of each side of the equation to solve for #x# while keeping the equation balanced. Remember, the square root of a number gives both a positive and negative result:

#sqrt(x^2) = +-sqrt(23/2)#

#x = +-sqrt(23/2)#

Or

#x = +-sqrt(11.5)#

Or

#x = +-3.391# rounded to the nearest thousandth

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