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Solve 1= (y+3)(2y-2)

Use foil on the right side.
1 = 2y^2 +6y -2y -6 so 1 = 2y^2 +4y -6

Now subtract 1 from boths sides giving us
0 = 2y^2 +4y -7 Now we need two numbers that multiply to be 2*(-7) or -14 and add to be 4. There doesn't appear to be any number so we will use the quadratic formula.

x =[ -b +/- sqr(b^2 -4ac)]/(2a)

Now a = 2, b = 4 and c = -7

x = [-4 +/- sqr(16+56)]/4

x = [-4 +/- sqr72]/4

x = [-4 +/- sqr(36*2)]/4 Factored the radicand

now 36 is a perfect square so we have

x = [-4 +/- 6*sqr2] /4 Now take a 2 out of every term giving us

x = [-2 +/- 3*sqr2]/2

So x = [-2 + 3* sqrt2]/2 or x = [-2 - 3*sqr2]/2

4 answers

Solve 1= (y+3)(2y-2)
Isn't the right side
2y^2 + 4y -6 and not the number you had for the y term?
Ok so now what?
So I looked at your answer and didn't see that what I wrote for the foil on the right is exactly what you have. I get the same final answer as you.
Thank you kind sir