450-125b^2+25b=b^2+b-6

Solution for 450-125b^2+25b=b^2+b-6 equation:

450-125b^2+25b=b^2+b-6

We move all terms to the left:

450-125b^2+25b-(b^2+b-6)=0

We get rid of parentheses

-125b^2-b^2+25b-b+6+450=0

We add all the numbers together, and all the variables

-126b^2+24b+456=0

a = -126; b = 24; c = +456;
Δ = b2-4ac
Δ = 242-4·(-126)·456
Δ = 230400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{230400}=480$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-480}{2*-126}=\frac{-504}{-252}
=+2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+480}{2*-126}=\frac{456}{-252}
=-1+17/21
$