If 2x^2 + 2x+3 =0 (you didn't mention =0. did you mean to?)
then use the quadratic formula as follows:
x=(-2±√[4-4(6)])/4
x=(-2±√[4-24])/4
x=(-2±√-20)/4
x=(-1±√-5)/2
use i (imaginary number=√-1)
x=(-1±i√5)/2
thats the answer.
By using the quadratic equation formula: x = -5 and x = 3
To solve for x: 2x2 + 10 = 58 2x2 = 48 x2 = 24 x = √24 x = √4 × √6 x = 2√6
2x2 + 5x - 3 (2x - 1)(x + 3)
2x2 + 7x + 3 = (2x + 1)(x + 3)If 2x2 + 7x + 3 = 0 then:(2x + 1)(x + 3) = 0⇒ (2x + 1) = 0 → x = -1/2or (x + 3) = 0 → x = - 3
x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3)
2x2+1x-3=0 4+x-3=0 x+1+0 x=-1 is your final answer
If that's + 2x2, the answer is x(x + 3)(x - 1) x = 0, -3, 1 If that's - 2x2, the answer is x(x - 3)(x + 1) x = 0, 3, -1
By using the quadratic equation formula: x = -5 and x = 3
To solve for x: 2x2 + 10 = 58 2x2 = 48 x2 = 24 x = √24 x = √4 × √6 x = 2√6
2x2 + 5x - 3 (2x - 1)(x + 3)
2x2 + 7x + 3 = (2x + 1)(x + 3)If 2x2 + 7x + 3 = 0 then:(2x + 1)(x + 3) = 0⇒ (2x + 1) = 0 → x = -1/2or (x + 3) = 0 → x = - 3
2x2-x-3 = 0= 4 - x - 3 = 0= 1 - x = 0= 1 = xTherefore x = 1
x3 + 2x2 + 3x + 6 = x2(x + 2) + 3(x + 2) = (x + 2)(x2 + 3)
2x2+5x+3 = (2x+3)(x+1) when factored
2x2+13x+15 = (2x+3)(x+5)
(2x + 1)(x + 3)
-2x3 + 2x2 + 12x =(-2x) (x2 - x - 6) =(-2x) (x+2) (x-3)