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2x^2+41x-252=0
a = 2; b = 41; c = -252;
Δ = b2-4ac
Δ = 412-4·2·(-252)
Δ = 3697
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{-(41)-\sqrt{3697}}{2*2}=\frac{-41-\sqrt{3697}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+\sqrt{3697}}{2*2}=\frac{-41+\sqrt{3697}}{4} $
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