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

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Answer 1 · 2016-07-07T16:35:56

$x=1/8 + sqrt145/8$ or $x=1/8 - sqrt145/8$

The Quadratic Formula is $(-b+-sqrt(b^2-4ac))/(2a)$

A Quadratic Equation is $ax^2 + bx + c = 0$

For the equation $-4x^2+x+9=0$

$a=-4$
$b=1$
$c=9$

Plug in the values

$(-1+-sqrt(1^2-4(-4)9))/(2(-4))$

$(-1+-sqrt(1+144))/(-8)$

$(-1+-sqrt(145))/(-8)$

$(-1)/-8 +-sqrt145/-8$

$1/8 + sqrt145/8$ or $1/8 - sqrt145/8$

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