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2x^2-4x-16=0
a = 2; b = -4; c = -16;
Δ = b2-4ac
Δ = -42-4·2·(-16)
Δ = 144
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*2}=\frac{-8}{4} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*2}=\frac{16}{4} =4 $
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