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2k^2-63+11k=0
a = 2; b = 11; c = -63;
Δ = b2-4ac
Δ = 112-4·2·(-63)
Δ = 625
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{625}=25$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-25}{2*2}=\frac{-36}{4} =-9 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+25}{2*2}=\frac{14}{4} =3+1/2 $
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