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.07x+.12x(2x+4400)=50850
We move all terms to the left:
.07x+.12x(2x+4400)-(50850)=0
We multiply parentheses
2x^2+.07x+4400x-50850=0
We add all the numbers together, and all the variables
2x^2+4400.07x-50850=0
a = 2; b = 4400.07; c = -50850;
Δ = b2-4ac
Δ = 4400.072-4·2·(-50850)
Δ = 19767416.0049
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{-(4400.07)-\sqrt{19767416.0049}}{2*2}=\frac{-4400.07-\sqrt{19767416.0049}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4400.07)+\sqrt{19767416.0049}}{2*2}=\frac{-4400.07+\sqrt{19767416.0049}}{4} $
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