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10b^2+17b+3=0
a = 10; b = 17; c = +3;
Δ = b2-4ac
Δ = 172-4·10·3
Δ = 169
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{169}=13$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-13}{2*10}=\frac{-30}{20} =-1+1/2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*10}=\frac{-4}{20} =-1/5 $
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