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3x^2-8x-16=0
a = 3; b = -8; c = -16;
Δ = b2-4ac
Δ = -82-4·3·(-16)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*3}=\frac{-8}{6} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*3}=\frac{24}{6} =4 $
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