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32d^2=98
We move all terms to the left:
32d^2-(98)=0
a = 32; b = 0; c = -98;
Δ = b2-4ac
Δ = 02-4·32·(-98)
Δ = 12544
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{12544}=112$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-112}{2*32}=\frac{-112}{64} =-1+3/4 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+112}{2*32}=\frac{112}{64} =1+3/4 $
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