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b^2+28b+192=0
a = 1; b = 28; c = +192;
Δ = b2-4ac
Δ = 282-4·1·192
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4}{2*1}=\frac{-32}{2} =-16 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4}{2*1}=\frac{-24}{2} =-12 $
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