Question: you solve and check this problem 30x2 – 45xy – 10x = .
I'm not too sure what it is that you want done here, but I've done some work on the problem, I hope that it helps.
30x^2 - 45xy - 10x = 0
30x^2 - 5(y+2)x = 0 -- multiply by 4 times the x^2-coefficient, i.e. 120
3600 - 600(9y+2)x = 0 -- now add and subtract the square of the x-coefficient, i.e. [5(9y+2)]^2.
3600 - 600(9y+2)x + 5^2(9y+2)^2 = 25(9y+2)^2 = 0 -- note that we have completed the square.
(60x - 5(9y+2))^2 = 25(9y+2)^2 -- take square root of both sides.
60x - 5(9y+2) = ± 5(9y+2)
if lhs = -5(9y+2), then x = 0 -- ignore this solution.
if lhs = +5(9y+2), then
60x = 10(9y+2)
6x = 9y + 2
y = (1/9)(6x - 2)
Answer: solutions to the original quadratic expression are any points lying on the line: y = (1/9)(6x - 2)