How do you solve #5r ^ { 2} - r + 7= 10#?

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Answer 1 · 2017-08-07T03:01:34

#r=+-1/10(1+sqrt61)#

#5r^2-r+7=10#

#5r^2-r-3=0#

This is a quadratic equation of the form: #ar^2+br+c=0#

Hence we can use the quadratic formula to solve for #r#:

#r=(-b+-sqrt(b^2-4ac))/(2a)# Where: #a=5, b=-1, c=-3#

#:. r = (-(-1)+-sqrt((-1)^2-4*5*(-3)))/(2*5)#

#= (1+-sqrt(1+60))/10#

#=1/10(1+-sqrt61)#

Since 61 is prime these roots cannot be simplified further.