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4x^2-4x-92=0
a = 4; b = -4; c = -92;
Δ = b2-4ac
Δ = -42-4·4·(-92)
Δ = 1488
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1488}=\sqrt{16*93}=\sqrt{16}*\sqrt{93}=4\sqrt{93}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{93}}{2*4}=\frac{4-4\sqrt{93}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{93}}{2*4}=\frac{4+4\sqrt{93}}{8} $
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