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(2w+4)w=160
We move all terms to the left:
(2w+4)w-(160)=0
We multiply parentheses
2w^2+4w-160=0
a = 2; b = 4; c = -160;
Δ = b2-4ac
Δ = 42-4·2·(-160)
Δ = 1296
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1296}=36$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-36}{2*2}=\frac{-40}{4} =-10 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+36}{2*2}=\frac{32}{4} =8 $
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