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2k^2+10=12
We move all terms to the left:
2k^2+10-(12)=0
We add all the numbers together, and all the variables
2k^2-2=0
a = 2; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·2·(-2)
Δ = 16
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{16}=4$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*2}=\frac{-4}{4} =-1 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*2}=\frac{4}{4} =1 $
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