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10b^2+61b+6=0
a = 10; b = 61; c = +6;
Δ = b2-4ac
Δ = 612-4·10·6
Δ = 3481
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{3481}=59$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(61)-59}{2*10}=\frac{-120}{20} =-6 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(61)+59}{2*10}=\frac{-2}{20} =-1/10 $
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