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-10w^2+15w=0
a = -10; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·(-10)·0
Δ = 225
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{225}=15$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*-10}=\frac{-30}{-20} =1+1/2 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*-10}=\frac{0}{-20} =0 $
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