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2w^2-15w+13=0
a = 2; b = -15; c = +13;
Δ = b2-4ac
Δ = -152-4·2·13
Δ = 121
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{121}=11$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-11}{2*2}=\frac{4}{4} =1 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+11}{2*2}=\frac{26}{4} =6+1/2 $
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