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5y^2+4y-12=0
a = 5; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·5·(-12)
Δ = 256
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{256}=16$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*5}=\frac{-20}{10} =-2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*5}=\frac{12}{10} =1+1/5 $
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