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25w^2+60w+32=0
a = 25; b = 60; c = +32;
Δ = b2-4ac
Δ = 602-4·25·32
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$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-20}{2*25}=\frac{-80}{50} =-1+3/5 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+20}{2*25}=\frac{-40}{50} =-4/5 $
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