Eavaluating the roots of quadratic equation If $b>a , c>0$ 
Determine the intervals that the roots of the equation
$(x-a)(x-b) -c =0$ belong to 
My work is to get the values of the roots in terms of a , b and c using the general form but i couldn't determine those intervals 
 A: First solve for $x$ the quadratic equation $(x-a)(x-b)-c=0$ by the general form we get that $x_1 =\frac{1}{2} \left(\sqrt{a^2-2 a b+b^2+4 c}+a+b\right)$ and $x_2=\frac{1}{2} \left(-\sqrt{a^2-2 a b+b^2+4 c}+a+b\right)$
substituting instead of $b$ the value $a$ we get that $x_1 >\frac{1}{2} \left(2 a+2 \sqrt{c}\right) $ and $x_2 > \frac{1}{2} \left(2 a-2 \sqrt{c}\right)$ because $b>a$.
substituting instead of $a$ the value $b$ we get that $x_1< \frac{1}{2} \left(2 b+2 \sqrt{c}\right)$ and $x_2 <\frac{1}{2} \left(2 b-2 \sqrt{c}\right) $
the smallest value of $\frac{1}{2} \left(2 a+2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 a-2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 b+2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 b-2 \sqrt{c}\right) $
is $\frac{1}{2} \left(2 a-2 \sqrt{c}\right)$ since $a<b$ and $c>0$.
the biggest value of $\frac{1}{2} \left(2 a+2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 a-2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 b+2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 b-2 \sqrt{c}\right) $
is $\frac{1}{2} \left(2 b+2 \sqrt{c}\right)$ since $b>a$ and $c>0$.
thus the roots are in the interval $[\frac{1}{2} \left(2 a-2 \sqrt{c}\right),\frac{1}{2} \left(2 b+2 \sqrt{c}\right)]$
A: The given equation is $x^2-(a+b)x+ab-c=0$. But $$\Delta=(a+b)^2-4(ab-c)=(a-b)^2+4c>0.$$ So the equation has two distinct real roots. The two roots are $\frac{(a+b)+\sqrt{(a-b)^2+4c}}{2}$ and $\frac{(a+b)-\sqrt{(a-b)^2+4c}}{2}$. Hence the roots are in the interval $[\frac{(a+b)-\sqrt{(a-b)^2+4c}}{2},\frac{(a+b)+\sqrt{(a-b)^2+4c}}{2}]$.  
