Question 916531
<pre>
Well, she tossed it all out there for you.  I'll go into
a little more detail, so you can see better how to use the
quadratic formula:

{{{2x^2+25x+33 = 0}}}

Compare to

{{{Ax^2+Bx+C=0}}}

and see that {{{A=2}}}, {{{B=25}}}, {{{C=33}}}

Then we use the quadratic formula which you must memorize:

{{{x = (-B +- sqrt( B^2-4AC ))/(2A) }}}

You might write empty parentheses for all the letters on the right,
like this:

{{{x = (-(""^("")) +- sqrt( (""^(""))^2-4(""^(""))(""^("")) ))/(2(""^(""))) }}}

There are TWO places to put the A's (A is 2)

{{{x = (-(""^("")) +- sqrt( (""^(""))^2-4(2^(""))(""^("")) ))/(2(2^(""))) }}}

There are TWO places to put the B's (B is 25)

{{{x = (-(25^("")) +- sqrt( (25^(""))^2-4(2^(""))(""^("")) ))/(2(2^(""))) }}}

But there is only ONE place to put the C (C is 33)

{{{x = (-(25^("")) +- sqrt( (25^(""))^2-4(2^(""))(33^("")) ))/(2(2^(""))) }}}

{{{x = (-25 +- sqrt(625-264)) /4 }}}

{{{x = (-25 +- sqrt(361)) /4 }}}

{{{x = (-25 +- 19) /4 }}}

Using the PLUS sign:

{{{x = (-25 + 19) /4 =(-6)/4=-3/2}}}

Using the MINUS sign:

{{{x = (-25 - 19) /4 =(-44)/4=-11}}}

Edwin</pre>