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9k^2-45=18k
We move all terms to the left:
9k^2-45-(18k)=0
a = 9; b = -18; c = -45;
Δ = b2-4ac
Δ = -182-4·9·(-45)
Δ = 1944
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}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1944}=\sqrt{324*6}=\sqrt{324}*\sqrt{6}=18\sqrt{6}$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18\sqrt{6}}{2*9}=\frac{18-18\sqrt{6}}{18} $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18\sqrt{6}}{2*9}=\frac{18+18\sqrt{6}}{18} $
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