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+50=5x+4.9x^2
We move all terms to the left:
+50-(5x+4.9x^2)=0
We get rid of parentheses
-4.9x^2-5x+50=0
a = -4.9; b = -5; c = +50;
Δ = b2-4ac
Δ = -52-4·(-4.9)·50
Δ = 1005
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1005}=\sqrt{1*1005}=\sqrt{1}*\sqrt{1005}=1\sqrt{1005}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1\sqrt{1005}}{2*-4.9}=\frac{5-1\sqrt{1005}}{-9.8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1\sqrt{1005}}{2*-4.9}=\frac{5+1\sqrt{1005}}{-9.8} $
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