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3k^2-25k=-42
We move all terms to the left:
3k^2-25k-(-42)=0
We add all the numbers together, and all the variables
3k^2-25k+42=0
a = 3; b = -25; c = +42;
Δ = b2-4ac
Δ = -252-4·3·42
Δ = 121
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{121}=11$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-11}{2*3}=\frac{14}{6} =2+1/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+11}{2*3}=\frac{36}{6} =6 $
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