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3x^2-20x-192=0
a = 3; b = -20; c = -192;
Δ = b2-4ac
Δ = -202-4·3·(-192)
Δ = 2704
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{2704}=52$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-52}{2*3}=\frac{-32}{6} =-5+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+52}{2*3}=\frac{72}{6} =12 $
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