Theory
Given quadratic equation in standard form:
[tex]ax^2+bx+c=0[/tex].
Its zeros (roots) are solutions that can be computed using quadratic formula:
[tex]
x_1=\dfrac{-b^2+\sqrt{b^2-4ac}}{2a} \\
x_2=\dfrac{-b^2-\sqrt{b^2-4ac}}{2a}
[/tex].
End Theory
Answer
We have quadratic equation:
[tex]3x^2+17-6=0[/tex].
Which can be simplified and rewritten to:
[tex]3x^2+0x+11=0[/tex].
From this we can extrapolate the coefficients:
[tex]a=3,b=0,c=11[/tex].
Insert the coefficients in the quadratic formulas and get calculate the roots (zeros/solutions):
[tex]x_1=\dfrac{-0^2+\sqrt{0^2-4(3)(11)}}{2\cdot3}\approx0+1.91i[/tex].
[tex]x_2=\dfrac{-0^2-\sqrt{0^2-4(3)(11)}}{2\cdot3}\approx0-0.91i[/tex].
So the solutions are two complex numbers.
End Answer
Hope this helps.
