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2x^2-12x-41=0
a = 2; b = -12; c = -41;
Δ = b2-4ac
Δ = -122-4·2·(-41)
Δ = 472
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{472}=\sqrt{4*118}=\sqrt{4}*\sqrt{118}=2\sqrt{118}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{118}}{2*2}=\frac{12-2\sqrt{118}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{118}}{2*2}=\frac{12+2\sqrt{118}}{4} $
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