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20x^2+16x-4=0
a = 20; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·20·(-4)
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24}{2*20}=\frac{-40}{40} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24}{2*20}=\frac{8}{40} =1/5 $
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