Question 1043584
<pre>
Apparently you made a mistake with the quadratic formula:

x = -3q<sup>2</sup>+150q 

substitute 150 for x

150 = -3q<sup>2</sup>+150q

Get 0 on the right side:

3q<sup>2</sup>-150q+150 = 0

To make it easier to solve, divide every 
term on both sides by 3 -- since you can!:

q<sup>2</sup>-50q+50 = 0

a=1, b=-50, c=50

{{{q = (-b +- sqrt( b^2-4ac ))/(2a) }}}

{{{q = (-(-50) +- sqrt((-50)^2-4(1)(50) ))/(2(1)) }}}

{{{q = (50 +- sqrt(2500-200))/2 }}}

{{{q = (50 +- sqrt(2300))/2 }}}

{{{q = (50 +- sqrt(100*23))/2 }}}

{{{q = (50 +- 10sqrt(23))/2 }}}

{{{q = (10(5 +- sqrt(23)))/2 }}}
 
{{{q = (""^5cross(10)(5 +- sqrt(23)))/cross(2) }}}

{{{q = 5(5 +- sqrt(23)) }}}

{{{q = 5(5 + sqrt(23)) }}} and {{{q = 5(5 - sqrt(23)) }}}

{{{q = 48.97915762}}} and {{{q = 1.020842383}}}

Both answer are correct, so give them both.  You should
try to find out where you went wrong, so that you won't 
make the same mistake in other problems.

Edwin</pre>