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2k^2-2=-6k
We move all terms to the left:
2k^2-2-(-6k)=0
We get rid of parentheses
2k^2+6k-2=0
a = 2; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·2·(-2)
Δ = 52
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{13}}{2*2}=\frac{-6-2\sqrt{13}}{4} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{13}}{2*2}=\frac{-6+2\sqrt{13}}{4} $
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