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5y^2+11y-12=0
a = 5; b = 11; c = -12;
Δ = b2-4ac
Δ = 112-4·5·(-12)
Δ = 361
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{361}=19$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*5}=\frac{-30}{10} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*5}=\frac{8}{10} =4/5 $
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