How do you solve #5x^2 - 2x +4 = 0# by quadratic formula?

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Answer 1 · 2017-07-12T01:00:49

#x=(1+isqrt19)/5,##x=(1-isqrt19)/5#

#5x^2-2x+4=0# is a quadratic equation in standard form: #ax^2+bx+c#, where #a=5#, #b=-2#, and #c=4#.

quadratic formula:

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

Substitute the values for #a,b, and c# into the formula.

#x=(-(-2)+-sqrt((-2)^2-4*5*4))/(2*5)#

Simplify.

#x=(2+-sqrt(4-80))/10#

#x=(2+-sqrt(-76))/10#

Prime factorize the number of the square root.

#x=(2+-isqrt(2xx2xx19))/10#

Simplify.

#x=(2+-2isqrt19)/10#

Reduce.

#x=(1+-isqrt19)/5#

Solutions for #x#.

#x=(1+isqrt19)/5,##x=(1-isqrt19)/5#