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2x^2-8x-46=0
a = 2; b = -8; c = -46;
Δ = b2-4ac
Δ = -82-4·2·(-46)
Δ = 432
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-12\sqrt{3}}{2*2}=\frac{8-12\sqrt{3}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+12\sqrt{3}}{2*2}=\frac{8+12\sqrt{3}}{4} $
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