0.12+b0.10(1+4b)=2.18

Solution for 0.12+b0.10(1+4b)=2.18 equation:

0.12+b0.10(1+4b)=2.18

We move all terms to the left:

0.12+b0.10(1+4b)-(2.18)=0

We add all the numbers together, and all the variables

b0.10(4b+1)+0.12-(2.18)=0

We add all the numbers together, and all the variables

b0.10(4b+1)-2.06=0

We multiply parentheses

4b^2+b-2.06=0

a = 4; b = 1; c = -2.06;
Δ = b2-4ac
Δ = 12-4·4·(-2.06)
Δ = 33.96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{33.96}}{2*4}=\frac{-1-\sqrt{33.96}}{8}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{33.96}}{2*4}=\frac{-1+\sqrt{33.96}}{8}
$