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49k^2-98k=0
a = 49; b = -98; c = 0;
Δ = b2-4ac
Δ = -982-4·49·0
Δ = 9604
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{9604}=98$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-98}{2*49}=\frac{0}{98} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+98}{2*49}=\frac{196}{98} =2 $
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