3x2+x+1=0
since it's not factorable, we use quadratic formula:
x = [-b +- sqrt(b^2 -4ac)]/(2a)
where a=3, b=1, and c=1
substituting:
x = [-1 +- sqrt(1^2 - 4*3*1)]/(2*3)
x = [-1 +- sqrt(-11)]/6
thus
x = [-1 + i*sqrt(11)]/6 ; and
x = [-1 - i*sqrt(-11)]/6
*note: i = sqrt(-1)
hope this helps~
oops, some typo:
x = [-1 + i*sqrt(11)]/6 ; and
x = [-1 - i*sqrt(11)]/6
To solve the equation 3x^2 + x + 1 = 0, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 3, b = 1, and c = 1.
Plugging these values into the quadratic formula, we have:
x = (-(1) ± √((1)^2 - 4(3)(1))) / (2(3))
Simplifying further:
x = (-1 ± √(1 - 12)) / 6
x = (-1 ± √(-11)) / 6
Since the square root of a negative number is not a real number, it means that this quadratic equation does not have any real solutions.