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2d^2+d-28=0
a = 2; b = 1; c = -28;
Δ = b2-4ac
Δ = 12-4·2·(-28)
Δ = 225
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{225}=15$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*2}=\frac{-16}{4} =-4 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*2}=\frac{14}{4} =3+1/2 $
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