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2w^2+15w-7=0
a = 2; b = 15; c = -7;
Δ = b2-4ac
Δ = 152-4·2·(-7)
Δ = 281
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{281}}{2*2}=\frac{-15-\sqrt{281}}{4} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{281}}{2*2}=\frac{-15+\sqrt{281}}{4} $
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