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k(k+1)-90=0
We multiply parentheses
k^2+k-90=0
a = 1; b = 1; c = -90;
Δ = b2-4ac
Δ = 12-4·1·(-90)
Δ = 361
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{361}=19$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*1}=\frac{-20}{2} =-10 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*1}=\frac{18}{2} =9 $
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