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-2w^2-3w+160=0
a = -2; b = -3; c = +160;
Δ = b2-4ac
Δ = -32-4·(-2)·160
Δ = 1289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{1289}}{2*-2}=\frac{3-\sqrt{1289}}{-4} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1289}}{2*-2}=\frac{3+\sqrt{1289}}{-4} $
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