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4q^2-31q-8=0
a = 4; b = -31; c = -8;
Δ = b2-4ac
Δ = -312-4·4·(-8)
Δ = 1089
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{1089}=33$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-33}{2*4}=\frac{-2}{8} =-1/4 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+33}{2*4}=\frac{64}{8} =8 $
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