(4/5)t-3=9

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Solution for (4/5)t-3=9 equation:



(4/5)t-3=9
We move all terms to the left:
(4/5)t-3-(9)=0
Domain of the equation: 5)t!=0
t!=0/1
t!=0
t∈R
We add all the numbers together, and all the variables
(+4/5)t-3-9=0
We add all the numbers together, and all the variables
(+4/5)t-12=0
We multiply parentheses
4t^2-12=0
a = 4; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·4·(-12)
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*4}=\frac{0-8\sqrt{3}}{8} =-\frac{8\sqrt{3}}{8} =-\sqrt{3} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*4}=\frac{0+8\sqrt{3}}{8} =\frac{8\sqrt{3}}{8} =\sqrt{3} $

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