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

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Answer 1 · 2016-07-14T09:30:08

$x=(-2+-2sqrt(13))/3$

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

solves equations of the type:

$ax^2+bx+c=0$

So, in this case, since

a=3; b=4: c=-3,

you have

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

$=(-4+-sqrt(16+36))/6$

$=(-4+-sqrt(52))/6$

$=(-4+-4sqrt(13))/6$

$=(-2+-2sqrt(13))/3$

How do you solve 3x^2+4x-3=03x^2+4x-3=0 using the quadratic formula?