Question 817230
start with what you are given and then write the related equations
let the numbers be x and y
x = 2y - 3
xy = 405
substitute for x in second equation
(2y - 3)y = 405
2y^2 -3y = 405
put equation in standard form
2y^2 -3y -405 = 0
use quadratic formula to solve for y
y = (3 + square root(3^2 -4*2*(-405)) / 2*2
y = (3 - square root(3^2 -4*2*(-405)) / 2*2
y = 15 or -13.5
solve for x in first equation
x = 27 or -30
check the answers
15*27 = 405
(-30)*(-13.5) = 405
our answers check, therefore there are two solution sets
x = 27 and y = 15
x = -30 and y = -13.5