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w^2+3w-700=0
a = 1; b = 3; c = -700;
Δ = b2-4ac
Δ = 32-4·1·(-700)
Δ = 2809
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{2809}=53$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-53}{2*1}=\frac{-56}{2} =-28 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+53}{2*1}=\frac{50}{2} =25 $
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