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2w^2-15w-13=0
a = 2; b = -15; c = -13;
Δ = b2-4ac
Δ = -152-4·2·(-13)
Δ = 329
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}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{329}}{2*2}=\frac{15-\sqrt{329}}{4} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{329}}{2*2}=\frac{15+\sqrt{329}}{4} $
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