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)):}#