(2/10)b=99

Solution for (2/10)b=99 equation:

(2/10)b=99

We move all terms to the left:

(2/10)b-(99)=0


Domain of the equation: 10)b!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

(+2/10)b-99=0

We multiply parentheses

2b^2-99=0

a = 2; b = 0; c = -99;
Δ = b2-4ac
Δ = 02-4·2·(-99)
Δ = 792
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:

$\sqrt{\Delta}=\sqrt{792}=\sqrt{36*22}=\sqrt{36}*\sqrt{22}=6\sqrt{22}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{22}}{2*2}=\frac{0-6\sqrt{22}}{4}
=-\frac{6\sqrt{22}}{4}
=-\frac{3\sqrt{22}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{22}}{2*2}=\frac{0+6\sqrt{22}}{4}
=\frac{6\sqrt{22}}{4}
=\frac{3\sqrt{22}}{2}
$