How do you find the roots, real and imaginary, of #y=-5x^2 +40x -34 # using the quadratic formula?

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Answer 1 · 2016-01-26T20:51:37

#4+-sqrt(9.2)#

#(-b+-sqrt(b^2-4*a*c))/(2*a)#

with a=-5, b=40 and c=-34 for this particular equation

#(-40+-sqrt(40^2-4*(-5)(-34)))/(2*(-5))#, which gives:

#(-40+-sqrt(1600-680))/(-10)#,

#(-40+-sqrt(920))/(-10)#,

#(40+-sqrt(920))/(10)#,

As 920 is not a perfect square, you can symplify the expression in several ways

#(40+-sqrt(4*230))/(10)=(20+-sqrt(230))/(5)=4+-sqrt(9.2)#