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20x^2-4x=20
We move all terms to the left:
20x^2-4x-(20)=0
a = 20; b = -4; c = -20;
Δ = b2-4ac
Δ = -42-4·20·(-20)
Δ = 1616
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1616}=\sqrt{16*101}=\sqrt{16}*\sqrt{101}=4\sqrt{101}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{101}}{2*20}=\frac{4-4\sqrt{101}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{101}}{2*20}=\frac{4+4\sqrt{101}}{40} $
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