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
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
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
3x2 + 30x + 48 is a quadratic polynomial. If 3x2 + 30x + 48 = 0, then: 3(x + 2)(x + 8) = 0 ⇒ x + 2 = 0 → x = -2 or x + 8 = 0 → x = -8
(3x-4)(x+5)
It is: x = 5/3 and x = -3
They are: x = 5/3 and x = -3
3x2-4x-15 = 0 (3x+5)(x-3) = 0 x = -5/3 or x = 3
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
Divide all terms by -1: 3x2+x -2 = 0 (3x-2)(x+1) = 0 x = 2/3 or x = -1
3x2+14x+8 = 0 x = -2/3 or x = -4
3x2 + 27x = 30 ∴ x2 + 9x - 10 = 0 ∴ (x + 10)(x - 1) = 0 ∴ x ∈ {-10, 1}