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2d^2+11d+14=0
a = 2; b = 11; c = +14;
Δ = b2-4ac
Δ = 112-4·2·14
Δ = 9
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{9}=3$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*2}=\frac{-14}{4} =-3+1/2 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*2}=\frac{-8}{4} =-2 $
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