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38z^2+49z=0
a = 38; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·38·0
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2401}=49$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*38}=\frac{-98}{76} =-1+11/38 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*38}=\frac{0}{76} =0 $
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