(2/5)b+1=11

Solution for (2/5)b+1=11 equation:

(2/5)b+1=11

We move all terms to the left:

(2/5)b+1-(11)=0


Domain of the equation: 5)b!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+2/5)b+1-11=0

We add all the numbers together, and all the variables

(+2/5)b-10=0

We multiply parentheses

2b^2-10=0

a = 2; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·2·(-10)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*2}=\frac{0-4\sqrt{5}}{4}
=-\frac{4\sqrt{5}}{4}
=-\sqrt{5}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*2}=\frac{0+4\sqrt{5}}{4}
=\frac{4\sqrt{5}}{4}
=\sqrt{5}
$