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49w^2+10w=0
a = 49; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·49·0
Δ = 100
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{100}=10$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*49}=\frac{-20}{98} =-10/49 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*49}=\frac{0}{98} =0 $
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