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4x^2-20x-16=0
a = 4; b = -20; c = -16;
Δ = b2-4ac
Δ = -202-4·4·(-16)
Δ = 656
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{41}}{2*4}=\frac{20-4\sqrt{41}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{41}}{2*4}=\frac{20+4\sqrt{41}}{8} $
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