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160x^2-155x-41=0
a = 160; b = -155; c = -41;
Δ = b2-4ac
Δ = -1552-4·160·(-41)
Δ = 50265
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{50265}=\sqrt{9*5585}=\sqrt{9}*\sqrt{5585}=3\sqrt{5585}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-155)-3\sqrt{5585}}{2*160}=\frac{155-3\sqrt{5585}}{320} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-155)+3\sqrt{5585}}{2*160}=\frac{155+3\sqrt{5585}}{320} $
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