How do you find the roots, real and imaginary, of $y= 3x^2-5x- 1$ using the quadratic formula?

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Answer 1 · 2016-01-29T08:02:24

$x=5/6+(sqrt3)/2,$ $5/6-(sqrt 3)/2$

$y=3x^2-5x-1$ is a quadratic equation in standard form, $y=ax+bx+c$ , where $a=3, b=-5, c=-1$ , and $y=0$ .

The roots are the solutions for $x$ .

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Substitute the known values into the equation.

$x=(-(-5)+-sqrt((-5)^2-(4*3*-1)))/(2*3)$

Simplify.

$x=(5+-sqrt(25+12))/6$

Add $25$ and $12$ .

$x=(5+-sqrt27)/6$

Simplify $sqrt(27)$ .

$sqrt27=sqrt(3*3*3)=sqrt(3^2*3)=3sqrt 3$

$x=(5+-3sqrt3)/6$

Write the value for $x$ as two fractions.

$x=5/6+-(3sqrt3)/6$

Simplify $3/6$ to $1/2$ .

$x=5/6+-(sqrt3)/2$

How do you find the roots, real and imaginary, of y= 3x^2-5x- 1 y= 3x^2-5x- 1 using the quadratic formula?