We determine the zeros of $y(x)=4x^2+8x+3$.
For this quadratic function it holds that $a=4$, $b=8$ and $c=3$. Hence, the discriminant is
$$D = b^2 - 4ac = 8^2-4\cdot4\cdot 3=16.$$
Since $16>0$, there are two solutions:
$x_1=\dfrac{-8+\sqrt{16}}{2\cdot 4}=-\frac{1}{2}$ and $x_2=\dfrac{-8-\sqrt{16}}{2\cdot 4}=-1\frac{1}{2}$.
For this quadratic function it holds that $a=4$, $b=8$ and $c=3$. Hence, the discriminant is
$$D = b^2 - 4ac = 8^2-4\cdot4\cdot 3=16.$$
Since $16>0$, there are two solutions:
$x_1=\dfrac{-8+\sqrt{16}}{2\cdot 4}=-\frac{1}{2}$ and $x_2=\dfrac{-8-\sqrt{16}}{2\cdot 4}=-1\frac{1}{2}$.