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48=2w(w+2)
We move all terms to the left:
48-(2w(w+2))=0
We calculate terms in parentheses: -(2w(w+2)), so:We get rid of parentheses
2w(w+2)
We multiply parentheses
2w^2+4w
Back to the equation:
-(2w^2+4w)
-2w^2-4w+48=0
a = -2; b = -4; c = +48;
Δ = b2-4ac
Δ = -42-4·(-2)·48
Δ = 400
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{400}=20$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*-2}=\frac{-16}{-4} =+4 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*-2}=\frac{24}{-4} =-6 $
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