If it was expressed as a quadratic equation it would have no roots because the discriminant is less than zero
2x2 + 3x - 20 = (x + 4)(2x - 5).
2x2 + 3x + 1 = (x + 1)(2x + 1)
d/dx 2x2+3x+7=4x+3
2x2 equals 5
-11
2X2 + 3X - 4 = 0the discriminant is,b2 - 4ac32 - 4(2)(- 4)9 + 32 = 4141 > 1this means there are two real roots to this equation
No.
If: 2x2-3x+1 = 0 Then: x = 1 or x = 1/2
2x2-4x-3x+6 2x2-7x+6 (2x+3)(x+2)
This expression could be reordered as 1 + 3x + 2x2.
No. You'd have to use the ABC-formula
(2x + 7)(x - 2)