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w^2+3w=130
We move all terms to the left:
w^2+3w-(130)=0
a = 1; b = 3; c = -130;
Δ = b2-4ac
Δ = 32-4·1·(-130)
Δ = 529
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{529}=23$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-23}{2*1}=\frac{-26}{2} =-13 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+23}{2*1}=\frac{20}{2} =10 $
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