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(2/3)k-3=9
We move all terms to the left:
(2/3)k-3-(9)=0
Domain of the equation: 3)k!=0We add all the numbers together, and all the variables
k!=0/1
k!=0
k∈R
(+2/3)k-3-9=0
We add all the numbers together, and all the variables
(+2/3)k-12=0
We multiply parentheses
2k^2-12=0
a = 2; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·2·(-12)
Δ = 96
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*2}=\frac{0-4\sqrt{6}}{4} =-\frac{4\sqrt{6}}{4} =-\sqrt{6} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*2}=\frac{0+4\sqrt{6}}{4} =\frac{4\sqrt{6}}{4} =\sqrt{6} $
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