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(8/3)x-5=-1
We move all terms to the left:
(8/3)x-5-(-1)=0
Domain of the equation: 3)x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
(+8/3)x-5-(-1)=0
We add all the numbers together, and all the variables
(+8/3)x-4=0
We multiply parentheses
8x^2-4=0
a = 8; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·8·(-4)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*8}=\frac{0-8\sqrt{2}}{16} =-\frac{8\sqrt{2}}{16} =-\frac{\sqrt{2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*8}=\frac{0+8\sqrt{2}}{16} =\frac{8\sqrt{2}}{16} =\frac{\sqrt{2}}{2} $
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