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16x^2+12x+2=0
a = 16; b = 12; c = +2;
Δ = b2-4ac
Δ = 122-4·16·2
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*16}=\frac{-16}{32} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*16}=\frac{-8}{32} =-1/4 $
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