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42x^2+41x+9=0
a = 42; b = 41; c = +9;
Δ = b2-4ac
Δ = 412-4·42·9
Δ = 169
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-13}{2*42}=\frac{-54}{84} =-9/14 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+13}{2*42}=\frac{-28}{84} =-1/3 $
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