Y²+10y+29=0. solve the quadratic equation
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Answer:
The answer for the question is imaginary
Step-by-step explanation:
Since,
We know that in an equation of ax^2+bx+c=0
By P= b^2-4ac,
When P>0 it have real roots
WhenP=0 have real and equal roots
When P<0 it have imaginary roots
So in the given equation,
a=1,b=10,c=29. Then b^2-4ac is equal to 100-116=-16
So this equal if solved wont get a real value
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