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2w^2+3w-629=0
a = 2; b = 3; c = -629;
Δ = b2-4ac
Δ = 32-4·2·(-629)
Δ = 5041
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{5041}=71$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-71}{2*2}=\frac{-74}{4} =-18+1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+71}{2*2}=\frac{68}{4} =17 $
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