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343x+4.9x^2=50
We move all terms to the left:
343x+4.9x^2-(50)=0
a = 4.9; b = 343; c = -50;
Δ = b2-4ac
Δ = 3432-4·4.9·(-50)
Δ = 118629
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{118629}=\sqrt{441*269}=\sqrt{441}*\sqrt{269}=21\sqrt{269}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(343)-21\sqrt{269}}{2*4.9}=\frac{-343-21\sqrt{269}}{9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(343)+21\sqrt{269}}{2*4.9}=\frac{-343+21\sqrt{269}}{9.8} $
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