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k^2+9k-99-99=0
We add all the numbers together, and all the variables
k^2+9k-198=0
a = 1; b = 9; c = -198;
Δ = b2-4ac
Δ = 92-4·1·(-198)
Δ = 873
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{873}=\sqrt{9*97}=\sqrt{9}*\sqrt{97}=3\sqrt{97}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{97}}{2*1}=\frac{-9-3\sqrt{97}}{2} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{97}}{2*1}=\frac{-9+3\sqrt{97}}{2} $
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