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20x^2-64x+12=0
a = 20; b = -64; c = +12;
Δ = b2-4ac
Δ = -642-4·20·12
Δ = 3136
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{3136}=56$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-56}{2*20}=\frac{8}{40} =1/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+56}{2*20}=\frac{120}{40} =3 $
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