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32n^2+22n+3=0
a = 32; b = 22; c = +3;
Δ = b2-4ac
Δ = 222-4·32·3
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-10}{2*32}=\frac{-32}{64} =-1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+10}{2*32}=\frac{-12}{64} =-3/16 $
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