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w^2+15w-700=0
a = 1; b = 15; c = -700;
Δ = b2-4ac
Δ = 152-4·1·(-700)
Δ = 3025
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{3025}=55$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-55}{2*1}=\frac{-70}{2} =-35 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+55}{2*1}=\frac{40}{2} =20 $
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