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11y^2+2y-15=0
a = 11; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·11·(-15)
Δ = 664
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{664}=\sqrt{4*166}=\sqrt{4}*\sqrt{166}=2\sqrt{166}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{166}}{2*11}=\frac{-2-2\sqrt{166}}{22} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{166}}{2*11}=\frac{-2+2\sqrt{166}}{22} $
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