Multiplied X in Quadratic Equation - Delta always lower than 0 I'm new to Quadratic Equations and I am following the two formulas to find $x_1$ and $x_2$ and they are:
$\Delta = b^2 - 4ac$
$x = \dfrac{-(b) \pm \sqrt\Delta}{2a}$
but delta always gives a negative number;
for example I have the following equation:
$6x^2 + 11x - 35 = 0$
What I did was:
1. Calculate the Delta ($\Delta$) by using the formula above
$11^2 - 4ac$
$11^2 = 121$ and  $4ac = -840 = 4\times6\times(-35)$
$\Delta = 121 - 840 = 719$
and got $-719$ as result
 A: You have e mistake in the sign:
$$
-4 ac = -4\cdot(6)\cdot(-35)=+840
$$
so:
$$
\Delta =b^2-4ac=b^2+(-4ac)= 121+840=961=31^2
$$
A: The discriminant $\Delta$ comes in three flavors.


*

*$\Delta > 0$: two real roots

*$\Delta = 0$: degenerate (repeated root)

*$\Delta < 0$: roots are complex conjugates $(x\pm iy)$




Here is a plot of the function in question:


This plot has two distinct, real roots. Therefore, $\Delta > 0$. 
To double check your work, note that
$$
  6 x^2+11 x-35 = (2 x+7) (3 x-5)
$$
so the roots are
$$
 x = -\frac{7}{2}, \qquad x = \frac{5}{3}
$$
A: $\Delta$ is not always negative, but sometimes it will be -- then the equation has no real solutions.
E.g.,:


*

*$x^2 - x + 1 = 0 \implies \Delta = 1^2-4 = -3 < 0$ has no real solutions

*$x^2 - 2x + 1 = 0 \implies \Delta = 2^2-4 = 0$ has exactly 1 real solution at $x=1$

*$2x^2 - 3x + 1 = 0 \implies \Delta = 3^2-4\times 2 = 1 > 0$ has 2 real solutions, $x=1$ and $x=1/2$.

A: The quadratic $6x^2 + 11x - 35$ has a=$6$, b=$11$ and c=$-35$. Therefore the discriminant $\Delta=b^2-4ac=(11)^2-4 \cdot 6 \cdot (-35)=121-(-840)=961$.
A: $6x^2+11x−35=0$
$a =6; b= 11; c= -35$
$b^2 = 121$.
$4ac = 4*6*(-35) =-840$
$b^2 - 4ac = 121 - (-840) = 121 + 841 = 961=31$.
So $\Delta = 961=31^2 > 0$ and $\sqrt{961} = 31$
$x = \frac {-11 \pm 31}{2*6}$.
$x = \frac {-42}{12}$ or $x = \frac {20}{12}$
$x = -\frac 72$ or $x = \frac 53$
A: Ok, let's start in dealing with your mistake and then talking more generally about what a negative descriminant ($\Delta$) means. 
In your case: $b^2 = 121$, $4ac=-840$ so $b^2-4ac=(121)-(-840)=+961$ (since minus a minus is plus). 
Now, there will be cases when your descriminant is negative, this means there are "no real solutions", you will learn about "non-real solutions" later but the case of a negative descriminant is easy to picture graphically. Think of solving a quadratic equation as looking for where a parabola crosses the x axis (finding the roots of the parabola), there will be parabolas that never cross the x axis, those are the cases when the descriminant is negative.
