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4x^2-12x-160=0
a = 4; b = -12; c = -160;
Δ = b2-4ac
Δ = -122-4·4·(-160)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-52}{2*4}=\frac{-40}{8} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+52}{2*4}=\frac{64}{8} =8 $
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