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5(w-4)3w-2=w+4
We move all terms to the left:
5(w-4)3w-2-(w+4)=0
We multiply parentheses
15w^2-60w-(w+4)-2=0
We get rid of parentheses
15w^2-60w-w-4-2=0
We add all the numbers together, and all the variables
15w^2-61w-6=0
a = 15; b = -61; c = -6;
Δ = b2-4ac
Δ = -612-4·15·(-6)
Δ = 4081
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{-(-61)-\sqrt{4081}}{2*15}=\frac{61-\sqrt{4081}}{30} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-61)+\sqrt{4081}}{2*15}=\frac{61+\sqrt{4081}}{30} $
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