Solve x² + x = 210.

Step-by-step solution

1. 1

   Read the structure before doing algebra. The left side is $x(x+1)$, the product of two consecutive integers. So the question is really: which consecutive integers multiply to $210$? Since $14\cdot 15=210$, expect $x=14$ — and a negative mirror solution, because $(-15)(-14)=210$ too.

2. 2

   Move everything to one side. A quadratic must be compared with zero before factoring:

   $x^2+x-210=0$

3. 3

   Find the factor pair. We need two numbers with product $-210$ and sum $+1$. Opposite signs are required (product negative), and the magnitudes must differ by $1$: $15$ and $-14$ fit, since

   $15\cdot(-14)=-210,\qquad 15+(-14)=1$

4. 4

   Factor and split.

   $x^2+x-210=(x+15)(x-14)=0\quad\Rightarrow\quad x+15=0\ \text{or}\ x-14=0$

5. 5

   Solve and check both roots.

   $x=-15\quad\text{or}\quad x=14$

   Check: $14^2+14=196+14=210$ ✓ and $(-15)^2+(-15)=225-15=210$ ✓. Both are valid; if the context required a positive quantity, only $x=14$ would be kept.