How do you complete the square to solve $2x^2 - 10x - 20 = 8$ ?

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Answer 1 · 2015-05-23T23:19:56

First notice that all the terms are divisible by $2$ , so let us divide both sides of the equation by $2$ to get:

$4=x^2-5x-10$

$= x^2-5x+25/4-25/4-10$

$= (x-5/2)^2-25/4-40/4$

$= (x-5/2)^2 - 65/4$

Add $65/4$ to both ends to get:

$(x-5/2)^2 = 4+65/4 = 16/4+65/4 = 81/4 = (9/2)^2$

So

$(x-5/2) = +-9/2$

Add $5/2$ to both sides to get:

$x = 5/2+-9/2$

That is $x = -4/2 = -2$ or $x = 14/2 = 7$ .

How do you complete the square to solve 2x^2 - 10x - 20 = 82x^2 - 10x - 20 = 8?