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12y^2+y-25=0
a = 12; b = 1; c = -25;
Δ = b2-4ac
Δ = 12-4·12·(-25)
Δ = 1201
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{-(1)-\sqrt{1201}}{2*12}=\frac{-1-\sqrt{1201}}{24} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1201}}{2*12}=\frac{-1+\sqrt{1201}}{24} $
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