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125x^2+25x-30=0
a = 125; b = 25; c = -30;
Δ = b2-4ac
Δ = 252-4·125·(-30)
Δ = 15625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{15625}=125$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-125}{2*125}=\frac{-150}{250} =-3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+125}{2*125}=\frac{100}{250} =2/5 $
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