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16x^2-12x-4=0
a = 16; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·16·(-4)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-20}{2*16}=\frac{-8}{32} =-1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+20}{2*16}=\frac{32}{32} =1 $
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