The equation can be re-written as $$\text{(x+10 5) 2(x-10 5) 2 -(x+10 5)(x-10 5) -x-1=0.}$$
We first prove that the equation has no negative roots.
Let $x\le 0.$ The equation above can be further re-arranged as $$\begin{align*}[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]=x-1.\end{align*}$$ The right hand side of the equation is negative. Therefore $$[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2)]<0,$$ and we have $-1<(x+10^5)(x-10^5) <2.$ Then the left hand side of the equation is bounded by $$|[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]|\le 3\times 3.$$ However, since $|(x+10^5)(x-10^5)|\le 2$ and $x<0,$ it follows that $|x+10^5| <\frac{2}{|x-10^5|}<2\times 10^{-5}$ for negative $x.$ Then $x<2\times 10^{-5}-10^5.$ The right hand side of the equation is then a large negative number. It cannot be equal to the left hand side which is bounded by 9.
Now let $x>0.$ When $x=10^5,$ the left hand side of equation (1) is negative. Therefore the equation has real roots on both side of $10^5$ , as its leading coefficient is positive. We will prove that $x=10^5$ is a good approximation of the roots (within $10^{-2}$ ). In fact, we can solve the "quadratic" equation (1) for $(x+10^5)(x-10^5)$ : $$(x+10^5)(x-10^5)=\frac{1\pm\sqrt{1+4(x+1)}}{2}.$$ Then $$x-10^5=\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}.$$ Easy to see that $|x-10^5| <1$ for positve $x.$ Therefore, $10^5-1<x<10^5+1.$ Then $$\begin{align*} |x-10^5|&=\left|\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\ &\le \left |\frac{1}{2(x+10^5)}\right |+\left |\frac{\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\ &\le \frac{1}{2(10^5-1+10^5)} +\frac{\sqrt{1+4(10^5+1+1)}}{2(10^5-1+10^5)} \\ &<10^{-2}. \end{align*}$$
Let $x_1$ be a root of the equation with $x_1<10^5.$ Then $0<10^5-x_1<10^{-2}$ and $$x_1-10^5=\frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}.$$ An aproximation of $x_1$ is defined as follows: $$\tilde{x}_1=10^5+\frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.$$ We check the error of the estimate: $$\begin{align*} |\tilde{x}_1-x_1|&=\left | \frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \right | \\ &\le \left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right |+\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |. \end{align*}$$
The first absolute value $$\left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right | =\frac{|x_1- 10^5|}{2(10^5+10^5)(x_1+10^5)}<10^{-12}.$$
The second absolute value $$\begin{align*} &\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)} - \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \right |\\ &\le \left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}\right |+\left |\frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |\\ &\le 10^{-7}+10^{-9}, \end{align*}$$ through a rationalized numerator.Therefore $|\tilde{x}_1-x_1|\le 10^{-6}.$
For a real root $x_2$ with $x_2>10^5,$ we choose $$\tilde{x}_2=10^5+\frac{1+\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.$$ We can similarly prove it has the desired approximation.