Answer:
[tex]x=\frac{5\±i√203}{6}[/tex](the lanky a is a square root)
Step-by-step explanation:
I suppose the equation is [tex]3x^2-5x+19=0[/tex]?
A good old quadratic equation! Back to the good old days when we are memorizing the formula...
[tex]x=\frac{-b\±√(b^2-4ac)}{2a}[/tex]
(Yes that little lanky a is a square root)
[tex]x=\frac{-(-5)\±√((-5)^2-4(3)(19))}{2(3)} \\x=\frac{5\±√(25-228)}{6}\\x=\frac{5\±√-203}{6} \\x=\frac{5\±i√203}{6}[/tex]
Since the discriminant is negative, there are no real solutions, but instead, we have complex solutions shown above.
