How to slove this problem in quadratic equation ? i cant seem to get the right answer, the back of the book says -3/2,1

2x² + x - 3 = 0. This is a quadratic equation of the form:
ax² + b . x + c = 0, where:

a = 2
b = 1
c = - 3

To find the roots (or zeros) of this quadratic equation we have to apply the following formula:

x' = [- b + √(b² - 4.a.c)]/2a  and x" = [- b - √(b² - 4.a.c)]/2a

Replace a , b and c by their respective values in both (x' and x")

x' = [-1 + √(1² - 4(2)(-3)]/2(2)  and x" = [-1 - √(1² - 4(2)(-3)]/2(2)

x' = [-1 + √(1+24)]/4   and  x' = [-1 - √(1+24)]/4

x' = (-1+5)/4    and x" = (-1 - 5)/4

x' = 4/4   and x" = -6/4

x' = 1   and x" = - 3/2

I’m going to infer you said parallel.solve for yy= -1/2x+7/4lines that are parallel have the same slope, different y intercept.so we know m is -1/2then from there solve for y intercept.