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x^2+14x-41=0
a = 1; b = 14; c = -41;
Δ = b2-4ac
Δ = 142-4·1·(-41)
Δ = 360
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-6\sqrt{10}}{2*1}=\frac{-14-6\sqrt{10}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+6\sqrt{10}}{2*1}=\frac{-14+6\sqrt{10}}{2} $
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