Well, you don't need the quadratic formula here. Note that $7x^2-3x = x(7x-3)$, so this is equal to $0$ only when either $x = 0$ or $7x-3 = 0$, that is, when $x=0$ or $x = \frac{3}{7}$.
0
Bumbble Comm
On
In your case $c = 0$.
$$
7 x^2 -3x = 7 x^2 -3x + 0 = a x^2 + bx + c
$$
So
$$
a = 7\quad b=-3\quad c=0
$$
0
Bumbble Comm
On
I think you mean $ax^{2}+bx+c=0$, here you simply have $c=0$. Further, there's no need for the quadratic formula, $x(7x-3)=0$
0
Bumbble Comm
On
HINT
Here $c=0$, so it is easier to simply factor to find the roots: $$x(7x-3)=0$$
0
user242594
On
$c$ is $0$ in this case. You can use the Quadratic formula right away to find $x=\frac{3\pm{\sqrt{9-0}}}{14}$ or factorize to get $f(x) = x(7x-3)$.
Well, you don't need the quadratic formula here. Note that $7x^2-3x = x(7x-3)$, so this is equal to $0$ only when either $x = 0$ or $7x-3 = 0$, that is, when $x=0$ or $x = \frac{3}{7}$.