I see something like this and I use the quadratic formula, which solves all quadratic equations. Discriminant, by inspection, looks good for two real roots.
X = -b +/- sqrt(b^2 - 4ac)/2a
a = -3
b = -1
c = 2
X = -(-1) +/- sqrt[(-1)^2 - 4(-3)(2)]/2(-3)
X = 1 +/- sqrt(25)/-6
X = {1 +/- 5}/-6
X = -1
X = 2/3
My TI-84 confirms these values for X
x=-.25.
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
3x2 + 2x = 16 ∴ 3x2 + 2x - 16 = 0 ∴ 3x2 - 6x + 8x - 16 = 0 ∴ 3x(x - 2) + 8(x - 2) = 0 ∴ (3x + 8)(x - 2) = 0 ∴ x ∈ {-8/3, 2}
3x2 + x - 2 = 0(3x - 2) (x + 1) = 0x = + 2/3x = -1
It is: 3x2+6x-11 = 0
x=-.25.
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
-3x2 + 22x = 0 You start by dividing both sides by x: ∴ -3x + 22 = 0 Then rearrange to solve for x: ∴3x = 22 ∴ x = 7 + 1/3
x2-5-4x2+3x = 0 -3x2+3x-5 = 0 or as 3x2-3x+5 = 0
3x2 + 2x = 16 ∴ 3x2 + 2x - 16 = 0 ∴ 3x2 - 6x + 8x - 16 = 0 ∴ 3x(x - 2) + 8(x - 2) = 0 ∴ (3x + 8)(x - 2) = 0 ∴ x ∈ {-8/3, 2}
3x2 + x - 2 = 0(3x - 2) (x + 1) = 0x = + 2/3x = -1
It is: 3x2+6x-11 = 0
56
(3x-4)(x+5)
2
3x2+14x+8 = 0 x = -2/3 or x = -4
3x2-9x = 0 x(3x-9) = 0 x = 0 or x = 3