How do you find the roots, real and imaginary, of $y=x^2 + 11x +28$ using the quadratic formula?

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Answer 1 · 2015-12-13T20:09:50

$x=-7,-4$

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

A quadratic equation is in the general form

$ax^2+bx+c$

So: ${(a=1),(b=11),(c=28):}$

Plug these into the quadratic formula.

$x=(-11+-sqrt(11^2-4(1)(28)))/(2(1))$

$x=(-11+-sqrt(121-112))/2$

$x=(-11+-sqrt9)/2$

$x=(-11+-3)/2$

${(x=(-11+3)/2=(-8)/2=color(blue)(-4)),(x=(-11-3)/2=(-14)/2=color(blue)(-7)):}$

How do you find the roots, real and imaginary, of y=x^2 + 11x +28 y=x^2 + 11x +28 using the quadratic formula?