3x2+2x-8=0
Delta = 22-4*3*-8= 100
So x = (-2 + sqrt(100))/(2*3) or x = (-2 - sqrt(100))/(2*3)
x = 4/3 or x=-2
3x2+x-4 = 0 (3x+4)(x-1) = 0 Solutions: x = 1 and x = -4/3 By using the quadratic equation formula.
Yes. (Assuming that -3x2 is the best representation of 3x2 that this browser will allow.)
3x2 - 10x - 8 = (3x + 2) (x - 4)
3x2 + 33x + 54 =3(x2 + 11x + 18) = 3(x + 9)(x + 2)
y2-3x2+6x+6y= 18 is in standard form. The vertex form would be (y+3)2/24 - (x-1)2/8 = 1
It has one double solution.
3x2=6+38=44
3x2 + 17x + c = 0, rearranging gives c = -3x2 - 17x
In general, no.
y2=x3+3x2
x=-.25.
No real roots.
Yes.
56
3
3x2 = 21 describes two single values, and does not have an x intercept. 3x2 = 21 ∴ x2 = 7 ∴ x = ±√7
no 3x2=6 then 4x-15= -60 so 6 added with - 60 = -54