How do you solve #4x^2+9x+5=0# using the quadratic formula?

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Answer 1 · 2016-05-13T13:34:13

#" "x= -10/8" and "-1#

Given that the standard form equation is: #y=ax^2+bx+c#

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

Given equation:#" "y=4x^2+9x+5#

So:#" "a=4" ; "b=9" ; "c=5#

Thus by substitution:

#" "x=(-9+-sqrt(9^2-4(4)(5)))/(2(4))#

#" "x=(-9+-sqrt(81-80))/(8)#

#" "x=(-9+-1)/8#

#" "x= -10/8" and "-1#