The quadratic function y=-x^2+10x-8 models the height of a trestle on a bridge. The x-axis represents ground level. To find where the section of the bridge meets ground level, solve 0=-x^2+10x-8 Where does this section of the bridge meet ground level? =-x^2+10x-8 Choose an equation that would be used to solve 0 x (x+25)^2=-8 (x-5)^2=17 x (x-10)^2=25 COMPLETE Solve the equation to find where the trestle meets ground level. Enter your answers from least to greatest and round to the nearest tenth. The trestle meets the ground at square units and square units.

To find where the section of the bridge meets ground level, we need to solve the equation $0=-x^{2}+10x-8$. <br /><br />Step 1: Rearrange the equation to standard form:<br />$-x^{2}+10x-8=0$<br /><br />Step 2: Use the quadratic formula to solve for x:<br />$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$<br /><br />In this case, a=-1, b=10, and c=-8. Plugging these values into the quadratic formula gives us:<br />$x=\frac{-10\pm\sqrt{10^2-4(-1)(-8)}}{2(-1)}$<br /><br />Simplifying further:<br />$x=\frac{-10\pm\sqrt{100-32}}{-2}$<br />$x=\frac{-10\pm\sqrt{68}}{-2}$<br />$x=\frac{-10\pm2\sqrt{17}}{-2}$<br />$x=5\pm\sqrt{17}$<br /><br />Therefore, the trestle meets the ground at $5-\sqrt{17}$ units and $5+\sqrt{17}$ units.