1/2b+2b+b+3b-5=270

Solution for 1/2b+2b+b+3b-5=270 equation:

1/2b+2b+b+3b-5=270

We move all terms to the left:

1/2b+2b+b+3b-5-(270)=0


Domain of the equation: 2b!=0

b!=0/2

b!=0

b∈R


We add all the numbers together, and all the variables

6b+1/2b-275=0

We multiply all the terms by the denominator


6b*2b-275*2b+1=0

Wy multiply elements

12b^2-550b+1=0

a = 12; b = -550; c = +1;
Δ = b2-4ac
Δ = -5502-4·12·1
Δ = 302452
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:

$\sqrt{\Delta}=\sqrt{302452}=\sqrt{4*75613}=\sqrt{4}*\sqrt{75613}=2\sqrt{75613}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-550)-2\sqrt{75613}}{2*12}=\frac{550-2\sqrt{75613}}{24}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-550)+2\sqrt{75613}}{2*12}=\frac{550+2\sqrt{75613}}{24}
$