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8q^2+23q-3=0
a = 8; b = 23; c = -3;
Δ = b2-4ac
Δ = 232-4·8·(-3)
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$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*8}=\frac{-48}{16} =-3 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*8}=\frac{2}{16} =1/8 $
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