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2y^2-55y+358=0
a = 2; b = -55; c = +358;
Δ = b2-4ac
Δ = -552-4·2·358
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-55)-\sqrt{161}}{2*2}=\frac{55-\sqrt{161}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-55)+\sqrt{161}}{2*2}=\frac{55+\sqrt{161}}{4} $
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