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(3k+2)(k-1)=0
We multiply parentheses ..
(+3k^2-3k+2k-2)=0
We get rid of parentheses
3k^2-3k+2k-2=0
We add all the numbers together, and all the variables
3k^2-1k-2=0
a = 3; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·3·(-2)
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$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*3}=\frac{-4}{6} =-2/3 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*3}=\frac{6}{6} =1 $
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