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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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-56}{2*20}=\frac{-120}{40} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+56}{2*20}=\frac{-8}{40} =-1/5 $
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