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-5x+20x^2=50
We move all terms to the left:
-5x+20x^2-(50)=0
a = 20; b = -5; c = -50;
Δ = b2-4ac
Δ = -52-4·20·(-50)
Δ = 4025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4025}=\sqrt{25*161}=\sqrt{25}*\sqrt{161}=5\sqrt{161}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{161}}{2*20}=\frac{5-5\sqrt{161}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{161}}{2*20}=\frac{5+5\sqrt{161}}{40} $
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