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16x^2+47x+3=0
a = 16; b = 47; c = +3;
Δ = b2-4ac
Δ = 472-4·16·3
Δ = 2017
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-\sqrt{2017}}{2*16}=\frac{-47-\sqrt{2017}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+\sqrt{2017}}{2*16}=\frac{-47+\sqrt{2017}}{32} $
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