How do you solve #4n^2 - 9n + 5 = 0#?

1 Answer
Jan 8, 2016

x=1 or 1.25

Explanation:

As we clearly see that factorisation is difficult so I suggest we use the quadratic formula for this.

The quadratic formula states that for an equation of the form #ax^2+bx+c=0#, the value of x can be calculated as

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

Comparing this with your question
x=n
a=4
b=-9
c=5

Plug in these values and find the values of n.