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12y^2+48y=0
a = 12; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·12·0
Δ = 2304
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{2304}=48$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*12}=\frac{-96}{24} =-4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*12}=\frac{0}{24} =0 $
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