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22d^2+49d=0
a = 22; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·22·0
Δ = 2401
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{2401}=49$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*22}=\frac{-98}{44} =-2+5/22 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*22}=\frac{0}{44} =0 $
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