Answer:
The answer in the procedure
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]0=-3x^{2} -2x+6[/tex]
so
[tex]a=-3\\b=-2\\c=6[/tex]
substitute in the formula
[tex]x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(-3)(6)}} {2(-3)}[/tex]
[tex]x=\frac{2(+/-)\sqrt{74}} {-6}[/tex]
[tex]x=\frac{-2-\sqrt{74}} {6}[/tex]
[tex]x=\frac{-2+\sqrt{74}} {6}[/tex]
Answer:
[tex]x=\frac{-(-2)\pm \sqrt{(-2)^2-4(-3)(6)}}{2(-3)}[/tex]
Step-by-step explanation:
The given quadratic equation is
[tex]-3x^2-2x+6=0[/tex] .... (1)
If a quadratic equation is defined as
[tex]ax^2+bx+c=0[/tex] .... (2)
then the quadratic formula is
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
On comparing (1) and (2), we get
[tex]a=-3,b=-2,c=6[/tex]
Substitute [tex]a=-3,b=-2,c=6[/tex] in the above formula.
[tex]x=\frac{-(-2)\pm \sqrt{(-2)^2-4(-3)(6)}}{2(-3)}[/tex]
Therefore, the correct substitution of the values a, b, and c in the quadratic formula is [tex]x=\frac{-(-2)\pm \sqrt{(-2)^2-4(-3)(6)}}{2(-3)}[/tex].
