Question

Find all complex solutions of 3x^2+3x+4=0.

Answer 1

Given equation is $3x^2 + 3x + 4 = 0$

Now we can compare it with general form of quadratic equation ($ax^2 + bx + c = 0$)
a = 3 , b = 3 and c = 4

Now we can apply quadratic formula which is given as
$x =\frac{ -b+/- \sqrt{b^2-4ac} }{2a}$

Now we can plugin value of a , b or c
$x = \frac{-3+/- \sqrt{(3)^2 - 4*3*4} }{2*3}$
         $= \frac{-3+/- \sqrt{9 - 48} }{6} = \frac{-3+/- \sqrt{-39} }{6}$
In general we know $\sqrt{-1} = i$
So we can write $\sqrt{-39 } = \sqrt{-1} * \sqrt{39} = i \sqrt{39}$
So
$x = \frac{-3+/-i \sqrt{39} }{6}$
So $x = \frac{-3+i \sqrt{39} }{6}$ or $x = \frac{-3- \sqrt{39} }{6}$