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16d^2+28d+12=0
a = 16; b = 28; c = +12;
Δ = b2-4ac
Δ = 282-4·16·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4}{2*16}=\frac{-32}{32} =-1 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4}{2*16}=\frac{-24}{32} =-3/4 $
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