SAT Question 2c3aefc9: Right Triangles And Trigonometry answer and explanation with Desmos solution

It’s given that in rectangle $PQRS$, the length of side $PS$ is $8$ more than the length of side $RS$. In a rectangle, opposite sides are congruent. Since side $QR$ is opposite side $PS$, it follows that side $QR$ is congruent to side $PS$. Thus, the length of side $QR$ is $8$ more than the length of side $RS$. Let $x$ represent the length of side $RS$. It follows that the length of side $QR$ is represented by the expression $x+8$. Since all angles in a rectangle are right angles, angle $R$ is a right angle. It follows that side $QR$, side $RS$, and line segment $QS$ form a right triangle, where line segment $QS$ is the hypotenuse and sides $QR$ and $RS$ are the legs. It’s given that the length of line segment $QS$ is $\sqrt{2,344}$. By the Pythagorean theorem, if a right triangle has a hypotenuse with length $c$ and legs with lengths $a$ and $b$, then $a^2+b^2=c^2$. Substituting $x$ for $a$, $x+8$ for $b$, and $\sqrt{2,344}$ for $c$ in this equation yields $x^2+(x+8{)}^2=(\sqrt{2,344}{)}^2$, or $x^2+(x+8{)}^2=2,344$. Expanding the left-hand side of this equation yields $x^2+x^2+16x+64=2,344$, or $2x^2+16x+64=2,344$. Dividing both sides of this equation by $2$ yields $x^2+8x+32=1,172$. To complete the square, subtracting $16$ from both sides of this equation yields $x^2+8x+16=1,156$. Since the left-hand side of this equation is a perfect square trinomial, it follows that $x^2+8x+16=1,156$ is equivalent to $(x+4{)}^2=1,156$. Taking the positive square root of both sides of this equation yields $x+4=34$. Subtracting $4$ from both sides of this equation yields $x=30$. Since the length of side $QR$ is represented by the expression $x+8$, it follows that the length of side $QR$ is $30+8$, or $38$.