(b+(3b-45)/2)+3b-45=400

Solution for (b+(3b-45)/2)+3b-45=400 equation:

(b+(3b-45)/2)+3b-45=400

We move all terms to the left:

(b+(3b-45)/2)+3b-45-(400)=0

We add all the numbers together, and all the variables

3b+(b+(3b-45)/2)-445=0

We multiply all the terms by the denominator


3b*2)+(b+(3b-45)-445*2)=0

We add all the numbers together, and all the variables

3b*2)+(b+(3b-45)=0

Wy multiply elements

6b^2+(3b-45)=0

We get rid of parentheses

6b^2+3b-45=0

a = 6; b = 3; c = -45;
Δ = b2-4ac
Δ = 32-4·6·(-45)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-33}{2*6}=\frac{-36}{12}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+33}{2*6}=\frac{30}{12}
=2+1/2
$