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6w^2+52w+96=0
a = 6; b = 52; c = +96;
Δ = b2-4ac
Δ = 522-4·6·96
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$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-20}{2*6}=\frac{-72}{12} =-6 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+20}{2*6}=\frac{-32}{12} =-2+2/3 $
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