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3/2k^2-16=80
We move all terms to the left:
3/2k^2-16-(80)=0
Domain of the equation: 2k^2!=0We add all the numbers together, and all the variables
k^2!=0/2
k^2!=√0
k!=0
k∈R
3/2k^2-96=0
We multiply all the terms by the denominator
-96*2k^2+3=0
Wy multiply elements
-192k^2+3=0
a = -192; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-192)·3
Δ = 2304
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*-192}=\frac{-48}{-384} =1/8 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*-192}=\frac{48}{-384} =-1/8 $
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