How do you solve $3x^2-4x-5=0$ using the quadratic formula?

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Answer 1 · 2018-04-05T05:38:12

$x=(2+sqrt19)/3,$ $(2-sqrt19)/3$

Solve:

$3x^2-4x-5=0$ is a quadratic equation in standard form:

$ax^2+bx+c$ ,

where:

$a=3$ , $b=-4$ , $c=-5$

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

Plug in the known values.

$x=(-(-4)+-sqrt((-4)^2-4*3*-5))/(2*3)$

Simplify.

$x=(4+-sqrt76)/6$

Prime factorize $76$ .

$x=(4+-sqrt(2^2xx19))/6$

Apply rule: $sqrt(a^2)=a$

$x=(4+-2sqrt19)/6$

Simplify.

$x=(2+-sqrt19)/3$

Solutions for $x$ .

$x=(2+sqrt19)/3,$ $(2-sqrt19)/3$

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