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2x(10x+1)=5
We move all terms to the left:
2x(10x+1)-(5)=0
We multiply parentheses
20x^2+2x-5=0
a = 20; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·20·(-5)
Δ = 404
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{101}}{2*20}=\frac{-2-2\sqrt{101}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{101}}{2*20}=\frac{-2+2\sqrt{101}}{40} $
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