Question 199172
The problem i am having is solving problems using the quadratic formula the the question is 3x squared equals negative 10x minus 4 thank you for any help you can provide.
<pre><font size = 4 color = "indigo"><b>

{{{3x^2=-10x-4}}}

We want to get 0 on the right, so we add {{{10x+4}}} 
to both sides:


{{{3x^2+10x+4=0}}}

The quadratic formula is stated this war:

If we have the equation:

{{{ax^2+bx+c=0}}}

then its solution is

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

In this equation, we have:

{{{3x^2+10x+4=0}}}

so {{{a=3}}}, {{{b=10}}}, {{{c=4}}}

{{{x = (-(10) +- sqrt((10)^2-4*(3)*(4) ))/(2*(3)) }}}

{{{x = (-10 +- sqrt(100-48 ))/6 }}}

{{{x = (-10 +- sqrt(52))/6 }}}

{{{x = (-10 +- sqrt(4*13))/6 }}}

{{{x = (-10 +- sqrt(4)sqrt(13))/6 }}}

{{{x = (-10 +- 2sqrt(13))/6 }}}

Factor {{{2}}} out of the top:

{{{x = (2(-5 +- sqrt(13)))/6 }}}

Cancel the 2 into the 6, getting 3:

{{{x = (cross(2)(-5 +- sqrt(13)))/(cross(6))[3] }}}

So the final answer is:

{{{x = (-5 +- sqrt(13))/3 }}}

which is really two separate solutions:

{{{x = (-5 + sqrt(13))/3 }}} and 
{{{x = (-5 - sqrt(13))/3 }}}

Edwin</pre>