# Quadratic formula errors

I'm clearly making a silly mistake here, but I can't see it.

EDIT: I missed brackets when typing out the expression to calculate. Apologies for timewasting.

I have the equation $(2x + 3)(5x + 1)=0$. Setting each bracket to $0$ I get my roots as $x=-\frac32$ and $x=-\frac15$.

To confirm my working I expand the brackets as $10x^2 + 17x + 3 = 0$. I then see my coefficients are $a=10$, $b=17$, and $c=3$.

Putting this into the formula I get $x = \frac{-17 \pm \sqrt{17^2 - 4\times10\times3}}{2\times10}$, which gives $x=-20$ and $x=-150$.

I really can't see what I've done wrong here. Any help appreciated. Thank you.

• It's hard to know where your mistake is if you don't show all your calculations... – DonAntonio May 23 '16 at 21:48
• I just typed it into google (-17 +sqrt(17**2 - 4*10*3)) / 2*10 EDIT: missed brackets on 2*10 – Ashiataka May 23 '16 at 21:56
• @As That doesn't help. The radical expression is 169, for example. Who knows what you got. – DonAntonio May 23 '16 at 22:33
• It was dividing by 2 then multiplying by 10 rather than dividing by 20. – Ashiataka May 23 '16 at 22:34

## 2 Answers

\begin{aligned} x_{1,2}&=\frac{-17\pm\sqrt{17^2-4\cdot 10\cdot 3}}{2\cdot 10}\\[6pt] &=\frac{-17\pm\sqrt{169}}{20}\\[6pt] &=\frac{-17\pm 13}{20}\\[6pt] &\implies x_{1}=\frac{-30}{20}=-\frac{3}{2},\quad x_{2}=\frac{-4}{20}=-\frac{1}{5} \end{aligned}

$$17^2-4\cdot10\cdot3\cdot=169=13^2\implies x_{1,2}=\frac{-17\pm13}{20}=-\frac32,\,-\frac15$$