10(2)=6(2)+b(2)

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

10(2)=6(2)+b(2)

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(+b^2+62)+102=0

We get rid of parentheses

-b^2-62+102=0

We add all the numbers together, and all the variables

-1b^2+40=0

a = -1; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-1)·40
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*-1}=\frac{0-4\sqrt{10}}{-2}
=-\frac{4\sqrt{10}}{-2}
=-\frac{2\sqrt{10}}{-1}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-1}=\frac{0+4\sqrt{10}}{-2}
=\frac{4\sqrt{10}}{-2}
=\frac{2\sqrt{10}}{-1}
$