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16x^2+47x-3=0
a = 16; b = 47; c = -3;
Δ = b2-4ac
Δ = 472-4·16·(-3)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2401}=49$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-49}{2*16}=\frac{-96}{32} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+49}{2*16}=\frac{2}{32} =1/16 $
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