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3x^2-36x-192=0
a = 3; b = -36; c = -192;
Δ = b2-4ac
Δ = -362-4·3·(-192)
Δ = 3600
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{3600}=60$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-60}{2*3}=\frac{-24}{6} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+60}{2*3}=\frac{96}{6} =16 $
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