These are the steps, with their explanations and conclusions:
1) Draw two triangles: ΔRSP and ΔQSP.
2) Since PS is perpendicular to the segment RQ, ∠ RSP and ∠ QSP are equal to 90° (congruent).
3) Since S is the midpoint of the segment RQ, the two segments RS and SQ are congruent.
4) The segment SP is common to both ΔRSP and Δ QSP.
5) You have shown that the two triangles have two pair of equal sides and their angles included also equal, which is the postulate SAS: triangles are congruent if any pair of corresponding sides and their included angles are equal in both triangles.
Then, now you conclude that, since the two triangles are congruent, every pair of corresponding sides are congruent, and so the segments RP and PQ are congruent, which means that the distance from P to R is the same distance from P to Q, i.e. P is equidistant from points R and Q
There is actually 2 ways to solve this, I will show you both.
The first is obvious, solve for x in the first one, and plug it into the 2nd one and get the answer
4x + 7 = 12
4x = 5
x = 
8() + 3
2 * 5 + 3
10 + 3
13
The 2nd option is to manipulate the 4x + 7 to be 8x + 3
4x + 7 = 12
start by moving the 7 over
4x = 5
multiply both sides by 2
8x = 10
and add 3 to both sides
8x + 3 = 13
X(x-1)(x^2+4)=(x^2-x)(x^2+4)=(x^4)+(4x^2)-(x^3)-(4x)=(x^4)-x(x^2+4)+(4x^2)
(0,0), (3,0), (-6,0), and (7,0)
I believe the GCF of those two is 12.
