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2w^2+10w-132=0
a = 2; b = 10; c = -132;
Δ = b2-4ac
Δ = 102-4·2·(-132)
Δ = 1156
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{1156}=34$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-34}{2*2}=\frac{-44}{4} =-11 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+34}{2*2}=\frac{24}{4} =6 $
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