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d^2+15d+14=0
a = 1; b = 15; c = +14;
Δ = b2-4ac
Δ = 152-4·1·14
Δ = 169
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{169}=13$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-13}{2*1}=\frac{-28}{2} =-14 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+13}{2*1}=\frac{-2}{2} =-1 $
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