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b^2+17b+42=0
a = 1; b = 17; c = +42;
Δ = b2-4ac
Δ = 172-4·1·42
Δ = 121
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{121}=11$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-11}{2*1}=\frac{-28}{2} =-14 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+11}{2*1}=\frac{-6}{2} =-3 $
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