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(5/6)k+2/3=4/3
We move all terms to the left:
(5/6)k+2/3-(4/3)=0
Domain of the equation: 6)k!=0We add all the numbers together, and all the variables
k!=0/1
k!=0
k∈R
(+5/6)k+2/3-(+4/3)=0
We multiply parentheses
5k^2+2/3-(+4/3)=0
We get rid of parentheses
5k^2+2/3-4/3=0
We multiply all the terms by the denominator
5k^2*3+2-4=0
We add all the numbers together, and all the variables
5k^2*3-2=0
Wy multiply elements
15k^2-2=0
a = 15; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·15·(-2)
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*15}=\frac{0-2\sqrt{30}}{30} =-\frac{2\sqrt{30}}{30} =-\frac{\sqrt{30}}{15} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*15}=\frac{0+2\sqrt{30}}{30} =\frac{2\sqrt{30}}{30} =\frac{\sqrt{30}}{15} $
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