2b^2-20b+100=68

Solution for 2b^2-20b+100=68 equation:

2b^2-20b+100=68

We move all terms to the left:

2b^2-20b+100-(68)=0

We add all the numbers together, and all the variables

2b^2-20b+32=0

a = 2; b = -20; c = +32;
Δ = b2-4ac
Δ = -202-4·2·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12}{2*2}=\frac{8}{4}
=2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12}{2*2}=\frac{32}{4}
=8
$