-2b(b-4)+4b=10

Solution for -2b(b-4)+4b=10 equation:

-2b(b-4)+4b=10

We move all terms to the left:

-2b(b-4)+4b-(10)=0

We add all the numbers together, and all the variables

4b-2b(b-4)-10=0

We multiply parentheses

-2b^2+4b+8b-10=0

We add all the numbers together, and all the variables

-2b^2+12b-10=0

a = -2; b = 12; c = -10;
Δ = b2-4ac
Δ = 122-4·(-2)·(-10)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*-2}=\frac{-20}{-4}
=+5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*-2}=\frac{-4}{-4}
=1
$