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0.2q^2-11q+30=0
a = 0.2; b = -11; c = +30;
Δ = b2-4ac
Δ = -112-4·0.2·30
Δ = 97
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{97}}{2*0.2}=\frac{11-\sqrt{97}}{0.4} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{97}}{2*0.2}=\frac{11+\sqrt{97}}{0.4} $
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