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y^2+11y-15.25=0
a = 1; b = 11; c = -15.25;
Δ = b2-4ac
Δ = 112-4·1·(-15.25)
Δ = 182
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{-(11)-\sqrt{182}}{2*1}=\frac{-11-\sqrt{182}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{182}}{2*1}=\frac{-11+\sqrt{182}}{2} $
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