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3d^2+22d+7=0
a = 3; b = 22; c = +7;
Δ = b2-4ac
Δ = 222-4·3·7
Δ = 400
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{400}=20$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-20}{2*3}=\frac{-42}{6} =-7 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+20}{2*3}=\frac{-2}{6} =-1/3 $
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