How do you find the zeros, real and imaginary, of #y=x^2-3x+298# using the quadratic formula?

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Answer 1 · 2016-03-09T20:58:19

#x=3/2+ (13sqrt(7))/2 i#

#x=3/2- (13sqrt(7))/2 i#

Standard form:#" "y=ax^2+bx+c#

Where:#" " x=(-b+-sqrt(b^2-4ac))/(2a)#

In your case:
#a=1#
#b=-3#
#c=298#

#=> x=(-(-3)+-sqrt((-3)^2-4(1)(298)))/(2(1))#

#=> x=(3+-sqrt(-1183))/2#

'~~~~~~~~~~~~~~~~~~~~~
#169xx7=1183#
#169=13^2#
'~~~~~~~~~~~~~~~~~~~
#=> x=(3+-sqrt(-1xx7xx13^2))/2#

#=> x=(3+-13sqrt(-1)sqrt7)/2#

#x=3/2+ (13sqrt(7))/2 i#

#x=3/2- (13sqrt(7))/2 i#