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25y^2+15y+2=0
a = 25; b = 15; c = +2;
Δ = b2-4ac
Δ = 152-4·25·2
Δ = 25
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{25}=5$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5}{2*25}=\frac{-20}{50} =-2/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5}{2*25}=\frac{-10}{50} =-1/5 $
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