Question 487465
<pre>
This problem could be done by factoring

3x² +  x - 4 = 0
 (3x+4)(x-1) = 0

3x+4= 0;  x-1=0
  3x=-4     x=1
   x={{{-4/3}}}

But you are to use the quadratic formula:

3x² +  x - 4 = 0   <--- take your problem and 

3x² + 1x - 4 = 0   <--- compare it to 

ax² + bx + c = 0   <--- this 

a = 3, b = 1, c = -4 which we substitute into:   


 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

 {{{x = (-(1) +- sqrt((1)^2-4*(3)*(-4) ))/(2*(3)) }}}

 {{{x = (-1 +- sqrt(1+48))/6 }}}

 {{{x = (-1 +- sqrt(49))/6 }}}

 {{{x = (-1 +- 7)/6 }}}

That breaks into two solutions:

 {{{x = (-1 + 7)/6 }}}  and  {{{x = (-1 - 7)/6 }}}

Simplifying the first one:

 {{{x = (-1 + 7)/6 }}}
 {{{x = 6/6 }}}
 {{{x = 1}}}

Simplifying the second one:

 {{{x = (-1 - 7)/6 }}}
 {{{x = -8/6 }}}
 {{{x = -4/3}}}

So the solutions are 1 and {{{-4/3}}}, the same
as when we solved it by factoring.  The reason we
have to use the quadratic formula is because most
quadratic equations will not factor.

Edwin</pre>