7. Which of the following represents the solution(s) to the following quadratic equation? x^2+14x-15=0

To solve the quadratic equation $x^{2}+14x-15=0$, we can use the quadratic formula:<br /><br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a$, $b$, and $c$ are the coefficients of the quadratic equation.<br /><br />In this case, $a = 1$, $b = 14$, and $c = -15$. Plugging these values into the quadratic formula, we get:<br /><br />$x = \frac{-14 \pm \sqrt{14^2 - 4(1)(-15)}}{2(1)}$<br /><br />Simplifying further, we have:<br /><br />$x = \frac{-14 \pm \sqrt{196 + 60}}{2}$<br /><br />$x = \frac{-14 \pm \sqrt{256}}{2}$<br /><br />$x = \frac{-14 \pm 16}{2}$<br /><br />This gives us two solutions:<br /><br />$x_1 = \frac{-14 + 16}{2} = 1$<br /><br />$x_2 = \frac{-14 - 16}{2} = -15$<br /><br />Therefore, the solutions to the quadratic equation $x^{2}+14x-15=0$ are $x = 1$ and $x = -15$.