How do I solve this quadratic equation? 2x^2=3x+5?

Question

I have to solve using zero product principle? How?

Answer 1

2x² = 3x + 5

2x² - 3x = 3x + 5 - 3x

2x² - 3x = 5

2x² - 3x - 5 = 5 - 5

2x² - 3x - 5 = 0

2x² + 2x - 5x - 5 = 0

2x(x + 1 - 5(x + 1) = 0

(2x- 5)(x + 1)

- - - - - - -

Roots

2x - 5 = 0

2x - 5 + 5 = 0 + 5

2x = 5

2x / 2 = 5 / 2

x = 5 / 2

x = 2 1/2

x = 2.5

- - - - - - -

Roots

x + 1 = 0

x + 1 - 1 = 0 - 1

x = - 1

- - - - - - -s-

Answer 2

2x^2 = 3x+5
bring everything to one side so you have a 0 to work with
meaning subtract 3x and 5 from both sides
2x^2 - 3x - 5 = 0
factor the left
(2x )(x ) = 0
(2x -5)(x+1) = 0
zero product principle, if two things multiplied equal zero then one of them has to be zero, therefore, set them both equal to zero to get your possible values of x that make that whole expression zero.
2x - 5 = 0 and x+1 = 0
x = 5/2 and x = -1

Answer 3

2x^2=3x+5
=>2x^2-3x-5 = 0
=> (2x-5)(x+1)=0
=> 2x-5 =0 or x+1=0
=> x = 2.5 or x= -1

Answer 4

get it all on one side (2x^2-3x-5)=0. then you have to break the part in the parentheses into F.O.I.L. form ex.(2x+1)(x-3)=0. Guess and check until it works and then solve for X.

Answer 5

- make it equal to 0 (2x^2-3x-5=0)
-use FOIL factor method (2x-5)(x+1)
x= -1 or 5/2

Answer 6

2x² - 3x - 5 = 0
(2x - 5).(x + 1) = 0
x = 5/2 , x = - 1