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12y^2+8y-20=0
a = 12; b = 8; c = -20;
Δ = b2-4ac
Δ = 82-4·12·(-20)
Δ = 1024
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{1024}=32$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-32}{2*12}=\frac{-40}{24} =-1+2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+32}{2*12}=\frac{24}{24} =1 $
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