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4y^2+49y+12=0
a = 4; b = 49; c = +12;
Δ = b2-4ac
Δ = 492-4·4·12
Δ = 2209
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{2209}=47$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-47}{2*4}=\frac{-96}{8} =-12 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+47}{2*4}=\frac{-2}{8} =-1/4 $
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