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5(k)=3k^2-7
We move all terms to the left:
5(k)-(3k^2-7)=0
We get rid of parentheses
-3k^2+5k+7=0
a = -3; b = 5; c = +7;
Δ = b2-4ac
Δ = 52-4·(-3)·7
Δ = 109
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{109}}{2*-3}=\frac{-5-\sqrt{109}}{-6} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{109}}{2*-3}=\frac{-5+\sqrt{109}}{-6} $
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