Solve Simultaneously: y=x^2-30 and x-2y=15

Answer is x=5 and y=-5
Don't know how that came to be and i dont know what method to use and on which values. (x,y or constant)

x- 2y = 15

=> 2y = x - 15

=> y =(x - 15)/ 2

Substitute for y:

(x - 15)/2 = x^2 - 30

=> x - 15 = 2x^2 - 60

=> 2x^2 - x - 45 = 0

Solve for x using quadratic formula:

x = -9/2 and x = 5

Substitute for x in x - 2y = 15:

x = -9/2 gives (-9/2) - 2y = 15, y = (-39/4)

x = 5, y = -5

Solutions: (5, -5) and (-9/2, -39/4)

Extra note: you gave us one of the solutions (5, -5) but since a quadratic is involved there will be two solutions unless the line touches the quadratic curve at it ape (which it doesn't).

y=x^2-30
x-2y=15

x = 2y + 15

y = (2y + 15)² - 30
y = 4y² + 60y + 15² - 30
4y² + 59y + 195 = 0

Δ = 59² - 4*4*195 = 361
y' = (- 59 + 19)/(2*4) = -5
y'' = (-59 - 19)/8 = -9.75

using the value of y', we have:
x - 2*(-5) = 15
x = 15 - 10
x = 5

y=x^2-30
x-2y=15

x-2(x^2-30)=15
-2x^2 + x +45 =0
2x^2 - x - 45 =0
(2x+9)(x-5)=0
x=-9/2 or x=5
y=39/2 or y=-5