Can someone help me solve this equation 2y^2=10y+7 And show how you got, it cuz I need help all...

drymer · 56-60, M Best Comment

OK, if you're going to use the quadratic formula then...

2y^2 -10y -7 = 0

so

a = 2
b = -10
c = -7

replace the a,b and c values in the quadratic formule and you'll get the value of Y

Makes sense?

SW-User

[b]2y^2 = 10y + 7
[/b]
This is a quadratic equation so you want to start by getting all the terms to one side so you have zero on the other...

[b]subtract 10y + 7 from both sides

2y^2 - 10y - 7 = 0[/b]

You can't factor this conveniently so you can either stick it into the quadratic formula or solve by completing the square...

[big]Completing the Square![/big]

First you want the coefficient of y^2 to be one...

[b]Divide both sides by 2

y^2 - 5y - 7/2 = 0[/b]

Now you want take half of the coefficient of y then square it and add and subtract it from the left side...

[b] Add and subtract (b/2)^2 = (5/2)^2 =25/4 from the left hand side

y^2 - 5y + (5/2)^2 - (5/2)^2 - 7/2 = 0
y^2 - 5y + (5/2)^2 - 25/4 - 7/2 = 0
y^2 - 5y + (5/2)^2 - 39/4 = 0[/b]

Now you can factor the first three term!!!!

[b](y - 5/2)^2 - 39/4 = 0[/b]

Now you want the squared thing to be on one side and everything else to the other...

[b]Add 39/4 to both sides

(y - 5/2)^2 = 39/4[/b]

Now you want to get rid of the square!

[b]take the square root of both sides

y - 5/2 = ±sqrt(39)/2[/b]

Now finally isolate y

[b]Add 5/2 to both sides

y = 5/2 ± sqrt(39)/2 [/b]

DONE!!!