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10b^2+28b+18=0
a = 10; b = 28; c = +18;
Δ = b2-4ac
Δ = 282-4·10·18
Δ = 64
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{64}=8$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8}{2*10}=\frac{-36}{20} =-1+4/5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8}{2*10}=\frac{-20}{20} =-1 $
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