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12w^2+17w-40=0
a = 12; b = 17; c = -40;
Δ = b2-4ac
Δ = 172-4·12·(-40)
Δ = 2209
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{2209}=47$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-47}{2*12}=\frac{-64}{24} =-2+2/3 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+47}{2*12}=\frac{30}{24} =1+1/4 $
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