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2q^2+11q-63=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:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{625}=25$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-25}{2*2}=\frac{-36}{4} =-9 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+25}{2*2}=\frac{14}{4} =3+1/2 $
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