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-0.25q^2+q+5=0
a = -0.25; b = 1; c = +5;
Δ = b2-4ac
Δ = 12-4·(-0.25)·5
Δ = 6
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{-(1)-\sqrt{6}}{2*-0.25}=\frac{-1-\sqrt{6}}{-0.5} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{6}}{2*-0.25}=\frac{-1+\sqrt{6}}{-0.5} $
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