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2d^2+14d-16=0
a = 2; b = 14; c = -16;
Δ = b2-4ac
Δ = 142-4·2·(-16)
Δ = 324
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{324}=18$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-18}{2*2}=\frac{-32}{4} =-8 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+18}{2*2}=\frac{4}{4} =1 $
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