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4q^2+39q-10=0
a = 4; b = 39; c = -10;
Δ = b2-4ac
Δ = 392-4·4·(-10)
Δ = 1681
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{1681}=41$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-41}{2*4}=\frac{-80}{8} =-10 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+41}{2*4}=\frac{2}{8} =1/4 $
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