-b^2+39=10b

Solution for -b^2+39=10b equation:

-b^2+39=10b

We move all terms to the left:

-b^2+39-(10b)=0

We add all the numbers together, and all the variables

-1b^2-10b+39=0

a = -1; b = -10; c = +39;
Δ = b2-4ac
Δ = -102-4·(-1)·39
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-16}{2*-1}=\frac{-6}{-2}
=+3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+16}{2*-1}=\frac{26}{-2}
=-13
$