Quadratic formula not working?

Question

Suppose I want to find the $ n $ for which

$$(n)(n+1)/2 = 10 \Longrightarrow n^2 + n - 20 = 0$$

Clearly a solution is $4$. But, suppose we wanted to find that solution by use of quadratic formula. We get

$$4 = \frac {-1 \pm \sqrt{1-80}}{2} $$

But that's clearly not true. What am I doing wrong?

Answer 1

You need a positive $$ \frac {-1 \pm \sqrt{1\color{green}{+}80}}{2} $$ Remember that the discriminant for $ax^2 +bx + c = 0$ is $b^2 -4ac$. Here $c = -20$. So $d = 1^2 - 4\cdot 1\cdot (-20) = 1 - (-80) = 1 + 80$.

Answer 2

$$\frac {-1 \pm \sqrt{1+80}}{2} $$

Answer 3

$$Ax^2+Bx+C=0\implies x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$$

$$\implies n=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(-20)}}{2\cdot1}$$

Answer 4

The discriminant is : $1 - 4(-20) = 81 = 9^2$ ! Not $-79$.

Answer 5

Let $\Delta = b^2 - 4ac$ be the discriminant. Then we see that $\Delta = (1)^2 - 4 (1)(-20)$, so $\Delta = 1 + 80 = 81$.

$$(-1) \times (-1) = 1$$