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Factor -y^2+13*y*b-2*b^2 squared

An expression to simplify:

The solution

You have entered [src]
   2               2
- y  + 13*y*b - 2*b 
$$- 2 b^{2} + \left(b 13 y - y^{2}\right)$$
-y^2 + (13*y)*b - 2*b^2
General simplification [src]
   2      2         
- y  - 2*b  + 13*b*y
$$- 2 b^{2} + 13 b y - y^{2}$$
-y^2 - 2*b^2 + 13*b*y
The perfect square
Let's highlight the perfect square of the square three-member
$$- 2 b^{2} + \left(b 13 y - y^{2}\right)$$
Let us write down the identical expression
$$- 2 b^{2} + \left(b 13 y - y^{2}\right) = \frac{161 y^{2}}{8} + \left(- 2 b^{2} + 13 b y - \frac{169 y^{2}}{8}\right)$$
or
$$- 2 b^{2} + \left(b 13 y - y^{2}\right) = \frac{161 y^{2}}{8} - \left(\sqrt{2} b - \frac{13 \sqrt{2} y}{4}\right)^{2}$$
Factorization [src]
/      /       _____\\ /      /       _____\\
|    y*\13 - \/ 161 /| |    y*\13 + \/ 161 /|
|b - ----------------|*|b - ----------------|
\           4        / \           4        /
$$\left(b - \frac{y \left(13 - \sqrt{161}\right)}{4}\right) \left(b - \frac{y \left(\sqrt{161} + 13\right)}{4}\right)$$
(b - y*(13 - sqrt(161))/4)*(b - y*(13 + sqrt(161))/4)
Powers [src]
   2      2         
- y  - 2*b  + 13*b*y
$$- 2 b^{2} + 13 b y - y^{2}$$
-y^2 - 2*b^2 + 13*b*y
Rational denominator [src]
   2      2         
- y  - 2*b  + 13*b*y
$$- 2 b^{2} + 13 b y - y^{2}$$
-y^2 - 2*b^2 + 13*b*y
Combining rational expressions [src]
     2                
- 2*b  + y*(-y + 13*b)
$$- 2 b^{2} + y \left(13 b - y\right)$$
-2*b^2 + y*(-y + 13*b)
Assemble expression [src]
   2      2         
- y  - 2*b  + 13*b*y
$$- 2 b^{2} + 13 b y - y^{2}$$
-y^2 - 2*b^2 + 13*b*y
Numerical answer [src]
-y^2 - 2.0*b^2 + 13.0*b*y
-y^2 - 2.0*b^2 + 13.0*b*y
Trigonometric part [src]
   2      2         
- y  - 2*b  + 13*b*y
$$- 2 b^{2} + 13 b y - y^{2}$$
-y^2 - 2*b^2 + 13*b*y
Common denominator [src]
   2      2         
- y  - 2*b  + 13*b*y
$$- 2 b^{2} + 13 b y - y^{2}$$
-y^2 - 2*b^2 + 13*b*y
Combinatorics [src]
   2      2         
- y  - 2*b  + 13*b*y
$$- 2 b^{2} + 13 b y - y^{2}$$
-y^2 - 2*b^2 + 13*b*y