Question 218354: solve the equation 2y^2-8y-5=0
you cant factor this out, i've tried to
im stuck on whats next, the home work says
1/2[2y^2-8y-5=0] then square both sides
why do you first use 1/2 with [2y^2-8y-5=0] ?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! solve the equation 2y^2-8y-5=0
you cant factor this out, i've tried to
im stuck on whats next, the home work says
1/2[2y^2-8y-5=0] then square both sides
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It might say multiply by 1/2, then complete the square of the eqn.
1/2[2y^2-8y-5=0]
--> y^2 - 4y = .25
Then add 4 (to both sides) to make the left side a square.
y^2 - 4y + 4 = 4.25
(y-2)^2 = 4.25
Now take sqrt of both sides
y-2 = sqrt(4.25)
y-2 = ~ 2.062
y = ~ 4.062
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y-2 = ~ -2.062
y = ~ 0.062
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