(3/2)b)+5=20-b

Solution for (3/2)b)+5=20-b equation:

(3/2)b)+5=20-b

We move all terms to the left:

(3/2)b)+5-(20-b)=0


Domain of the equation: 2)b)+5-(20!=0

b∈R


We add all the numbers together, and all the variables

(+3/2)b)+5-(-1b+20)=0

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


-1b*2)b)+5-(+(+3=0

Wy multiply elements

-2b^2+3=0

a = -2; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-2)·3
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-2}=\frac{0-2\sqrt{6}}{-4}
=-\frac{2\sqrt{6}}{-4}
=-\frac{\sqrt{6}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-2}=\frac{0+2\sqrt{6}}{-4}
=\frac{2\sqrt{6}}{-4}
=\frac{\sqrt{6}}{-2}
$