Find the zero(s) for y=4x^2+20x+25 -2.5 3.2 -4.1 -1.6

To find the zeros of the quadratic function $y=4x^{2}+20x+25$, we need to solve the equation $4x^{2}+20x+25=0$.<br /><br />We can use the quadratic formula to solve this equation. The quadratic formula is given by:<br /><br />$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$<br /><br />In this case, $a=4$, $b=20$, and $c=25$. Plugging these values into the quadratic formula, we get:<br /><br />$x=\frac{-20\pm\sqrt{20^2-4(4)(25)}}{2(4)}$<br /><br />Simplifying further, we have:<br /><br />$x=\frac{-20\pm\sqrt{400-400}}{8}$<br /><br />$x=\frac{-20\pm\sqrt{0}}{8}$<br /><br />$x=\frac{-20}{8}$<br /><br />$x=-2.5$<br /><br />Therefore, the zero of the quadratic function $y=4x^{2}+20x+25$ is $x=-2.5$.