(2/5)b+6=10

Solution for (2/5)b+6=10 equation:

(2/5)b+6=10

We move all terms to the left:

(2/5)b+6-(10)=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+6-10=0

We add all the numbers together, and all the variables

(+2/5)b-4=0

We multiply parentheses

2b^2-4=0

a = 2; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·2·(-4)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*2}=\frac{0-4\sqrt{2}}{4}
=-\frac{4\sqrt{2}}{4}
=-\sqrt{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*2}=\frac{0+4\sqrt{2}}{4}
=\frac{4\sqrt{2}}{4}
=\sqrt{2}
$