Verified answer

2y + 3 = x

y = x²

x = +/- sqrt y

2y + 3 = +/- sqrt y

4y^2 + 5y + 9 = 0

y

= [-5 +/- sqrt (25 - 144)]/8

= (-5 +/- i sqrt 119)/8

x = sqrt [(-5 +/- i sqrt 119)/8]

If you substitute you get

2x^2 + 3 = x

or 2x^2 - x + 3 = 0

Solving you get 2 complex solutions

[1±sqrt(-23)]/4 or (1/4)±sqrt(23)i/4

dnadan1

y = (2y+3)^2 = 4y^2+12y +9

4y^2+11y +9=0

y = (-11 +/- sqrt[121-4 (4) (9) ])/8 = (-11+/- sqrt[-23])/8

you are correct , no real solution

if this leads to a quadratic equation then I'm on the right track.. if it doesn't then sorry but idk wtf im looking at.

Put y = x² in first equation, you will get

2x²-x+3 = 0

as in the given quadratic equation

b^2-4ac< 0

hence no real solution.