2x^2-5x=3 how do you solve by quadratic equation?

Question

Im not trying to cheat i need the help, its not a test.

Answer 1

2x^2 - 5x - 3 =0
2x^2 - 6x + x - 3 = 0
2x(x-3) + (x-3) = 0
(2x+1)(x-3) = 0
so ANS. = x= -1/2 x= 3

There are 3 ways of doing this :
(i)Factorising (as shown above)
(ii)Completing the square
(iii)Using the Quadratic formula (derived from (ii)

Answer 2

Since you asked about solving by the "quadratic equation," it seems that you are asking for help in learning how to use the Quadratic Formula. The theorem states this: The equation ax^2 + bx + c = 0, with a NE 0, has roots
x = (-b +/- sqrt(b^2 - 4ac))/2a.

Your equation is 2x^2 - 5x - 3 = 0, so a = 2, b = -5, and
c = -3. Therefore x = (-(-5) +/- sqrt((-5)^2 - 4(2)(-3)))/2(2) =
(5 +/- sqrt(25 + 24))/4 = (5 +/- 7)/4 = (5 + 7)/4 and (5 - 7)/4,
so the roots are x = 3 and x = -1/2.