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0 = 2x^2 + 2x - 4 = 2(x^2 + x - 2) = 2(x-1)(x+2) Thus, there are two real solutions: x = 1 and x = -2
2x - y = 8 x + y = 1 These are your two equations. They will have two solutions since you have two variables. The solutions are x=3 and y=-2
(2x + 1)(2x + 1) or (2x + 1)2 so x = -0.5
Only one: (3,-2)
I messed up the question I ment what are the solutions to 2x^2 + 5x + 8 = 6