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125k^2-45=0
a = 125; b = 0; c = -45;
Δ = b2-4ac
Δ = 02-4·125·(-45)
Δ = 22500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{22500}=150$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-150}{2*125}=\frac{-150}{250} =-3/5 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+150}{2*125}=\frac{150}{250} =3/5 $
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