-(6/5)b+(1/10)=(17/2)

Solution for -(6/5)b+(1/10)=(17/2) equation:

-(6/5)b+(1/10)=(17/2)

We move all terms to the left:

-(6/5)b+(1/10)-((17/2))=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

-(+6/5)b+(+1/10)-((+17/2))=0

We multiply parentheses

-6b^2+(+1/10)-((+17/2))=0

We get rid of parentheses

-6b^2+1/10-((+17/2))=0

We calculate fractions


-6b^2+()/()+()/()=0

We add all the numbers together, and all the variables

-6b^2+2=0

a = -6; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-6)·2
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-6}=\frac{0-4\sqrt{3}}{-12}
=-\frac{4\sqrt{3}}{-12}
=-\frac{\sqrt{3}}{-3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-6}=\frac{0+4\sqrt{3}}{-12}
=\frac{4\sqrt{3}}{-12}
=\frac{\sqrt{3}}{-3}
$