How do you find the roots, real and imaginary, of $y= x^2 + 4x - 1$ using the quadratic formula?

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Answer 1 · 2016-07-29T22:15:55

${-2-sqrt(5),-2+sqrt(5)}$

The quadratic formula states that given a quadratic equation

$ax^2+bx+c=0$

we have

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

Then, with $x^2+4x-1=0$ , we have $a=1, b=4, c=-1$ . Applying the formula gives us

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

$=(-4+-sqrt(20))/2$

$=(-4+-2sqrt(5))/2$

$=-2+-sqrt(5)$

Thus the two roots of the given equation are $-2+sqrt(5)$ and $-2-sqrt(5)$

How do you find the roots, real and imaginary, of y= x^2 + 4x - 1 y= x^2 + 4x - 1 using the quadratic formula?