1 Answer

Zartch · Answer 1 · 2021-12-16T16:41:06+0000

Theory

Given quadratic equation in standard form:

ax^2+bx+c=0 .

Its zeros (roots) are solutions that can be computed using quadratic formula:

$x_1=(-b^2+√(b^2-4ac))/(2a) \\x_2=(-b^2-√(b^2-4ac))/(2a) $ .

End Theory

Answer

We have quadratic equation:

3x^2+17-6=0 .

Which can be simplified and rewritten to:

3x^2+0x+11=0 .

From this we can extrapolate the coefficients:

a=3,b=0,c=11 .

Insert the coefficients in the quadratic formulas and get calculate the roots (zeros/solutions):

$x_1=(-0^2+√(0^2-4(3)(11)))/(2\cdot3)\approx0+1.91i$ .

$x_2=(-0^2-√(0^2-4(3)(11)))/(2\cdot3)\approx0-0.91i$ .

So the solutions are two complex numbers.

End Answer

Hope this helps.

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