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2/5k+16=3/10k+13
We move all terms to the left:
2/5k+16-(3/10k+13)=0
Domain of the equation: 5k!=0
k!=0/5
k!=0
k∈R
Domain of the equation: 10k+13)!=0We get rid of parentheses
k∈R
2/5k-3/10k-13+16=0
We calculate fractions
20k/50k^2+(-15k)/50k^2-13+16=0
We add all the numbers together, and all the variables
20k/50k^2+(-15k)/50k^2+3=0
We multiply all the terms by the denominator
20k+(-15k)+3*50k^2=0
Wy multiply elements
150k^2+20k+(-15k)=0
We get rid of parentheses
150k^2+20k-15k=0
We add all the numbers together, and all the variables
150k^2+5k=0
a = 150; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·150·0
Δ = 25
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{25}=5$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*150}=\frac{-10}{300} =-1/30 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*150}=\frac{0}{300} =0 $
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