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3x^2-16x-12=0
a = 3; b = -16; c = -12;
Δ = b2-4ac
Δ = -162-4·3·(-12)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-20}{2*3}=\frac{-4}{6} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+20}{2*3}=\frac{36}{6} =6 $
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