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0.01x^2=5x+49=0
We move all terms to the left:
0.01x^2-(5x+49)=0
We get rid of parentheses
0.01x^2-5x-49=0
a = 0.01; b = -5; c = -49;
Δ = b2-4ac
Δ = -52-4·0.01·(-49)
Δ = 26.96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{26.96}}{2*0.01}=\frac{5-\sqrt{26.96}}{0.02} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{26.96}}{2*0.01}=\frac{5+\sqrt{26.96}}{0.02} $
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