How do you find the roots, real and imaginary, of $y=2x^2 -12x -5$ using the quadratic formula?

Question

1775 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-07T17:01:08

There are two real roots that are irrational: $x=3 pm 1/2 sqrt(46)$

The quadratic formula says the roots of the function $y=ax^2+bx+c$ are $x=(-b pm sqrt(b^{2}-4ac))/(2a)$ .

For the present problem, $a=2$ , $b=-12$ , and $c=-5$ . Hence, the roots are

$(12 pm sqrt{144-4*2*(-5)})/(2*2)=3 pm 1/4 sqrt{184}$

$=3 pm 1/4 sqrt{4}sqrt{46}=3 pm 1/2 sqrt(46)$ .

How do you find the roots, real and imaginary, of y=2x^2 -12x -5y=2x^2 -12x -5 using the quadratic formula?