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3x^2-4x-160=0
a = 3; b = -4; c = -160;
Δ = b2-4ac
Δ = -42-4·3·(-160)
Δ = 1936
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{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-44}{2*3}=\frac{-40}{6} =-6+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+44}{2*3}=\frac{48}{6} =8 $
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