Solve 2x^2-3y-5=0?

Question

can u solve Solve 2x^2-3y-5=0 by compleating the square. and can you please show me the steps thanks

Answer 1

it cant be -3y
that should be -3x
2x^2 -3x -5 = 0

factor number2 out for two first term
2(x^2 -(3/2)x +............ ) -5 =0
the number behind +..... will be (-3/2)/2 then square
(-3/2)/2 = (-3/4) square that to get 9/16
2(x^2 -(3/2)x+ 9/16) -5 = 9/8
because 2 *(9/16) = 9/8 if you add on theleft hand by 9/8 you have to add 9/8 to the right hand side
add 5 both sides
2 { x- (3/4)}^2 = 9/8 +5
2{ x- (3/4)}^2 = 49/8
divide by 2
(x-(3/4))^2 = 49/16
take square root both sides
√(x-(3/4))^2) = ± √(49/16)
x -(3/4) = ± 7/4

we got 2 answers
x-(3/4) = 7/4 and x-(3/4) = -7/4
x = 10/4 and x = -1

or
x = 5/2 and x = -1

Answer 2

I believe that you mean 2x^2 - 3x - 5 = 0. Is not it?
So: 0 = 2x^2 - 3x - 5
= 2x^2 - 3x + 9/8 - 9/8 - 5
= 2 (x^2 - 3x/2 + 9/16) - (5 + 9/8)
= 2 (x^2 - 2x(3/4) + (3/4)^2) - (5 + 9/8)
= 2(x - 3/4)^2 - (49/8)
or: (x - 3/4)^2 = (49/16) = (7/4)^2
Thus: x - 3/4 = ± 7/4
You can do the rest.