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

Question

7472 views around the world

You can reuse this answer
Creative Commons License

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:

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

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

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

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

How do you solve 4x^2+9x+5=04x^2+9x+5=0 using the quadratic formula?