I will use the quadratic equation here. Try discriminant. b^2 -4ac
1^2 - 4(-3)(5) = 61 is > 1, so two real roots
X = -1 +/- sqrt[1^2 - 4(-3)(5)]/2(1)
-1 +/- sqrt(61)/2
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x3 + 3x2 - 9x + 5 = 0 has roots of -5,1 and 1. CHECK : x3 + 3x2 - 9x + 5 = (x + 5)(x - 1)(x - 1)
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
If you mean: 3x2+8+5 = 0 Then it crosses the x axis at points -1 and -5/3
(3x-4)(x+5)
If you meant 3x2 - 5x + 2, here is your answer 3x2 - 5x +2 = 0 => 3x2 - 3x - 2x + 2 = 0 => 3x(x-1) - 2(x-1) = 0 => (3x - 2) (x-1) = 0 => x = 2/3 or, x = 1