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3d^2+11d+10=0
a = 3; b = 11; c = +10;
Δ = b2-4ac
Δ = 112-4·3·10
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-1}{2*3}=\frac{-12}{6} =-2 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+1}{2*3}=\frac{-10}{6} =-1+2/3 $
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