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2d^2-13d+14=0
a = 2; b = -13; c = +14;
Δ = b2-4ac
Δ = -132-4·2·14
Δ = 57
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}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{57}}{2*2}=\frac{13-\sqrt{57}}{4} $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{57}}{2*2}=\frac{13+\sqrt{57}}{4} $
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