Quartic polynomials?

$|(x-3)(x-4)|=\lambda$

$(x-3)(x-4)=\lambda \; \vee \; (x-3)(x-4)=-\lambda$

$x^2-7x+12-\lambda=0 \; \vee \; x^2-7x+12+\lambda=0$

All solution are: $x=\frac{7 \pm \sqrt{1\pm4\lambda}}{2}$

Since we want to have a double root, one of the two radicands must be zero. We also have the condition $\lambda \geq 0$ , therefore must be $\lambda=\frac{1}{4}=\boxed{0.25}$

The three solutions are $\{\frac{7-\sqrt{2}}{2},\;\frac{7}{2},\;\frac{7+\sqrt{2}} {2}\}$