# Solve using Quadratic Formula Question, What went wrong?

Okay the question is to solve using quadratic formula and the question is $10r^2-6=0$ and after using the $-b \pm \frac{\sqrt{ b^2-4ac}}{2a}$ I got $-0 \pm 0^2-\frac{\sqrt{240}}{20}$...and I don't know where to go after that...i thought about taking square root but than i get $15$ something $20$??

• Well, is there not a simpler way to solve the question? $10r^2=6$... – Feyre Sep 8 '15 at 21:27
• As @Feyre said, it is simple solution. $r=\pm \sqrt{\frac 35}$ – user249332 Sep 8 '15 at 21:29
• Just write out the prime factorization of $240$. You will see that it can be factored as $16\times 15$ – John Joy Sep 8 '15 at 22:48

$\pm \frac{\sqrt{240}}{20} = \pm\frac{\sqrt{240}}{\sqrt{400}} =\pm \sqrt{\frac{240}{400}} = \pm\sqrt{\frac{3}{5}}$
• @MATHASKER: $\frac{\sqrt{15}}5 = \frac{\sqrt{15}}{\sqrt{25}} = \sqrt{\frac{15}{25}} = \sqrt{\frac 35}$ – Peter Phipps Sep 8 '15 at 23:11
• Most math classes will require reductions of all radicals, fractions, and rationalized denominators. So the reduction you'd see would be: $\frac {\sqrt{240}}{20} = \frac {\sqrt{16*15}}{20} = \frac {4 \sqrt{15}} {4*5} = \frac {\sqrt{15}}{5}$. – Daniel R. Collins Sep 9 '15 at 3:59