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20x^2-26x+8=0
a = 20; b = -26; c = +8;
Δ = b2-4ac
Δ = -262-4·20·8
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-6}{2*20}=\frac{20}{40} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+6}{2*20}=\frac{32}{40} =4/5 $
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