b-6=2(b^2-3)

Solution for b-6=2(b^2-3) equation:

b-6=2(b^2-3)

We move all terms to the left:

b-6-(2(b^2-3))=0


We calculate terms in parentheses: -(2(b^2-3)), so:

2(b^2-3)

We multiply parentheses

2b^2-6

Back to the equation:

-(2b^2-6)


We get rid of parentheses

-2b^2+b+6-6=0

We add all the numbers together, and all the variables

-2b^2+b=0

a = -2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-2)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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