(2/3)s-(5/6)s+0.5=-3/2

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Solution for (2/3)s-(5/6)s+0.5=-3/2 equation:



(2/3)s-(5/6)s+0.5=-3/2
We move all terms to the left:
(2/3)s-(5/6)s+0.5-(-3/2)=0
Domain of the equation: 3)s!=0
s!=0/1
s!=0
s∈R
Domain of the equation: 6)s!=0
s!=0/1
s!=0
s∈R
We add all the numbers together, and all the variables
(+2/3)s-(+5/6)s+0.5-(-3/2)=0
We multiply parentheses
2s^2-5s^2+0.5-(-3/2)=0
We get rid of parentheses
2s^2-5s^2+0.5+3/2=0
We multiply all the terms by the denominator
2s^2*2-5s^2*2+3+(0.5)*2=0
We add all the numbers together, and all the variables
2s^2*2-5s^2*2+4=0
Wy multiply elements
4s^2-10s^2+4=0
We add all the numbers together, and all the variables
-6s^2+4=0
a = -6; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-6)·4
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-6}=\frac{0-4\sqrt{6}}{-12} =-\frac{4\sqrt{6}}{-12} =-\frac{\sqrt{6}}{-3} $
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-6}=\frac{0+4\sqrt{6}}{-12} =\frac{4\sqrt{6}}{-12} =\frac{\sqrt{6}}{-3} $

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