(b/2)+(2b/3)=(35/2)

Solution for (b/2)+(2b/3)=(35/2) equation:

(b/2)+(2b/3)=(35/2)

We move all terms to the left:

(b/2)+(2b/3)-((35/2))=0

We add all the numbers together, and all the variables

(+b/2)+(+2b/3)-((+35/2))=0

We get rid of parentheses

b/2+2b/3-((+35/2))=0

We calculate fractions


16b^2/()+3b/()+()/()=0

We add all the numbers together, and all the variables

16b^2/()+3b/()+1=0

We multiply all the terms by the denominator


16b^2+3b+1*()=0

We add all the numbers together, and all the variables

16b^2+3b=0

a = 16; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·16·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*16}=\frac{-6}{32}
=-3/16
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*16}=\frac{0}{32}
=0
$