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k^2=11k-30
We move all terms to the left:
k^2-(11k-30)=0
We get rid of parentheses
k^2-11k+30=0
a = 1; b = -11; c = +30;
Δ = b2-4ac
Δ = -112-4·1·30
Δ = 1
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{1}=1$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2*1}=\frac{10}{2} =5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2*1}=\frac{12}{2} =6 $
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