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23k^2-21k=0
a = 23; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·23·0
Δ = 441
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{441}=21$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*23}=\frac{0}{46} =0 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*23}=\frac{42}{46} =21/23 $
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