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3(7k+3)(7k-3)=0
We use the square of the difference formula
49k^2-9=0
a = 49; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·49·(-9)
Δ = 1764
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{1764}=42$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42}{2*49}=\frac{-42}{98} =-3/7 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42}{2*49}=\frac{42}{98} =3/7 $
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