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12x^2+32x+16=0
a = 12; b = 32; c = +16;
Δ = b2-4ac
Δ = 322-4·12·16
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16}{2*12}=\frac{-48}{24} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*12}=\frac{-16}{24} =-2/3 $
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