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20x+12x^2=0
a = 12; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·12·0
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*12}=\frac{-40}{24} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*12}=\frac{0}{24} =0 $
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