How do you find the zeros, real and imaginary, of # y = 7x^2 +12x + 2 # using the quadratic formula?

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Answer 1 · 2017-01-17T07:16:39

The solutions are #S={(-6+sqrt22)/7,(-6-sqrt22)/7}#

We compare this equation

#7x^2+12x+2=0#

to

#ax^2+b+c=0#

We start by calculating the discriminant

#Delta=b^2-4ac=12^2-4*7*2=144-56=88#

Therefore,

As #Delta>0#, there are 2 real roots

#x=(-b+-sqrtDelta)/(2a)#

#x_1=(-12+sqrt88)/14=(-6+sqrt22)/7#

#x_2=(-12-sqrt88)/14=(-6-sqrt22)/7#