SAT Question e11294f9: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Subtracting $7$ from both sides of the equation $x^2+6x+7=0$ yields $x^2+6x=-7$. To complete the square, adding ${\left(\frac{6}{2}\right)}^2$, or $3^2$, to both sides of this equation yields $x^2+6x+3^2=-7+3^2$, or ${\left(x+3\right)}^2=2$. Taking the square root of both sides of this equation yields $x+3=\pm \sqrt{2}$. Subtracting $3$ from both sides of this equation yields $x=-3\pm \sqrt{2}$. Therefore, the solutions $r$ and $s$ to the equation $x^2+6x+7=0$ are $-3-\sqrt{2}$ and $-3+\sqrt{2}$. Since $r<s$, it follows that $r=-3-\sqrt{2}$ and $s=-3+\sqrt{2}$. Subtracting $8$ from both sides of the equation $x^2+8x+8=0$ yields $x^2+8x=-8$. To complete the square, adding ${\left(\frac{8}{2}\right)}^2$, or $4^2$, to both sides of this equation yields $x^2+8x+4^2=-8+4^2$, or ${\left(x+4\right)}^2=8$. Taking the square root of both sides of this equation yields $x+4=\pm \sqrt{8}$, or $x+4=\pm 2\sqrt{2}$. Subtracting $4$ from both sides of this equation yields $x=-4\pm 2\sqrt{2}$. Therefore, the solutions $t$ and $u$ to the equation $x^2+8x+8=0$ are $-4-2\sqrt{2}$ and $-4+2\sqrt{2}$. Since $t<u$, it follows that $t=-4-2\sqrt{2}$ and $u=-4+2\sqrt{2}$. It's given that the solutions to $x^2+14x+c=0$, where $c$ is a constant, are $r+t$ and $s+u$. It follows that this equation can be written as $\left(x-\left(r+t\right)\right)\left(x-\left(s+u\right)\right)=0$, which is equivalent to $x^2-\left(r+t+s+u\right)x+\left(r+t\right)\left(s+u\right)=0$. Therefore, the value of $c$ is $\left(r+t\right)\left(s+u\right)$. Substituting $-3-\sqrt{2}$ for $r$, $-4-2\sqrt{2}$ for $t$, $-3+\sqrt{2}$ for $s$, and $-4+2\sqrt{2}$ for $u$ in this equation yields $\left(\left(-3-\sqrt{2}\right)+\left(-4-2\sqrt{2}\right)\right)\left(\left(-3+\sqrt{2}\right)+\left(-4+2\sqrt{2}\right)\right)$, which is equivalent to $\left(-7-3\sqrt{2}\right)\left(-7+3\sqrt{2}\right)$, or $\left(-7\right)\left(-7\right)-\left(3\sqrt{2}\right)\left(3\sqrt{2}\right)$, which is equivalent to $49-18$, or $31$. Therefore, the value of $c$ is $31$.