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12y^2+40y+25=0
a = 12; b = 40; c = +25;
Δ = b2-4ac
Δ = 402-4·12·25
Δ = 400
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{400}=20$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20}{2*12}=\frac{-60}{24} =-2+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20}{2*12}=\frac{-20}{24} =-5/6 $
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