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133=5w+2w^2
We move all terms to the left:
133-(5w+2w^2)=0
We get rid of parentheses
-2w^2-5w+133=0
a = -2; b = -5; c = +133;
Δ = b2-4ac
Δ = -52-4·(-2)·133
Δ = 1089
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{1089}=33$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-33}{2*-2}=\frac{-28}{-4} =+7 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+33}{2*-2}=\frac{38}{-4} =-9+1/2 $
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