How do you solve #2x^2 - 3x + 4 = 0#?

1 Answer
Gió
May 16, 2015

This is a quadratic equation of the form:
#ax^2+bx+c=0#
Where:
#a=2#
#b=-3#
#c=4#
So you can use the Quadratic Formula:
#x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)#
#=(3+-sqrt(9-32))/4#
The negative argument of the square root tells you that you will not get Real solutions (you'll get Complex Solutions).
#x_(1,2)=(3+-sqrt(-1*23))/4=(3+-isqrt(23))/4#
Where #i=sqrt(-1)#

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