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22y^2-3y-10=0
a = 22; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·22·(-10)
Δ = 889
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{-(-3)-\sqrt{889}}{2*22}=\frac{3-\sqrt{889}}{44} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{889}}{2*22}=\frac{3+\sqrt{889}}{44} $
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