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k^2-20k+36=0
a = 1; b = -20; c = +36;
Δ = b2-4ac
Δ = -202-4·1·36
Δ = 256
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{256}=16$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16}{2*1}=\frac{4}{2} =2 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16}{2*1}=\frac{36}{2} =18 $
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