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20x^2+8x-28=0
a = 20; b = 8; c = -28;
Δ = b2-4ac
Δ = 82-4·20·(-28)
Δ = 2304
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{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-48}{2*20}=\frac{-56}{40} =-1+2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+48}{2*20}=\frac{40}{40} =1 $
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