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m^2+169m=0
a = 1; b = 169; c = 0;
Δ = b2-4ac
Δ = 1692-4·1·0
Δ = 28561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{28561}=169$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(169)-169}{2*1}=\frac{-338}{2} =-169 $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(169)+169}{2*1}=\frac{0}{2} =0 $
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