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 solve this problem, we need to find the points where the trestle meets the ground level. Let's assume the equation of the trestle is given by a quadratic function in the form of $y = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants.<br /><br />To find the points where the trestle meets the ground level, we need to set $y$ equal to 0 and solve for $x$. This will give us the x-coordinates of the points where the trestle meets the ground.<br /><br />The equation becomes:<br />$0 = ax^2 + bx + c$<br /><br />We can solve this quadratic equation using the quadratic formula:<br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />Let's assume the coefficients are $a = 1$, $b = -4$, and $c = 3$. Plugging these values into the quadratic formula, we get:<br />$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)}$<br />$x = \frac{4 \pm \sqrt{16 - 12}}{2}$<br />$x = \frac{4 \pm \sqrt{4}}{2}$<br />$x = \frac{4 \pm 2}{2}$<br /><br />So, the two solutions are:<br />$x_1 = \frac{4 + 2}{2} = 3$<br />$x_2 = \frac{4 - 2}{2} = 1$<br /><br />Therefore, the trestle meets the ground at 1 unit and 3 units.<br /><br />The trestle meets the ground at $\boxed{1}$ units and $\boxed{3}$ units.