x^2-14x+40=0 lesser x= greater x=

To solve the quadratic equation $x^{2}-14x+40=0$, we can use the quadratic formula:<br /><br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a = 1$, $b = -14$, and $c = 40$.<br /><br />Plugging in the values, we get:<br /><br />$x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(40)}}{2(1)}$<br /><br />Simplifying further:<br /><br />$x = \frac{14 \pm \sqrt{196 - 160}}{2}$<br /><br />$x = \frac{14 \pm \sqrt{36}}{2}$<br /><br />$x = \frac{14 \pm 6}{2}$<br /><br />So, the solutions are:<br /><br />$x_1 = \frac{14 + 6}{2} = 10$<br /><br />$x_2 = \frac{14 - 6}{2} = 4$<br /><br />Therefore, the lesser value of $x$ is $4$ and the greater value of $x$ is $10$.