SAT Question 158591f0: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Applying the distributive property to the left-hand side of the given equation, $x\left(x+1\right)-56$, yields $x^2+x-56$. Applying the distributive property to the right-hand side of the given equation, $4x\left(x-7\right)$, yields $4x^2-28x$. Thus, the equation becomes $x^2+x-56=4x^2-28x$. Combining like terms on the left- and right-hand sides of this equation yields $0=\left(4x^2-x^2\right)+\left(-28x-x\right)+56$, or $3x^2-29x+56=0$. For a quadratic equation in the form $a{x}^2+bx+c=0$, where $a$, $b$, and $c$ are constants, the quadratic formula gives the solutions to the equation in the form $x=\frac{\left(-b\pm \sqrt{b^2-4ac}\right)}{2a}$. Substituting $3$ for $a$, $-29$ for $b$, and $56$ for $c$ from the equation $3x^2-29x+56=0$ into the quadratic formula yields $x=\frac{\left(29\pm \sqrt{{\left(-29\right)}^2-4\left(3\right)\left(56\right)}\right)}{2\left(3\right)}$, or $x=\frac{29}{6}\pm \frac{13}{6}$. It follows that the solutions to the given equation are $\frac{29}{6}+\frac{13}{6}$ and $\frac{29}{6}-\frac{13}{6}$. Adding these two solutions gives the sum of the solutions: $\frac{29}{6}+\frac{13}{6}+\frac{29}{6}-\frac{13}{6}$, which is equivalent to $\frac{29}{6}+\frac{29}{6}$, or $\frac{29}{3}$. Note that 29/3, 9.666, and 9.667 are examples of ways to enter a correct answer.