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25y^2+15y=2
We move all terms to the left:
25y^2+15y-(2)=0
a = 25; b = 15; c = -2;
Δ = b2-4ac
Δ = 152-4·25·(-2)
Δ = 425
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-5\sqrt{17}}{2*25}=\frac{-15-5\sqrt{17}}{50} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+5\sqrt{17}}{2*25}=\frac{-15+5\sqrt{17}}{50} $
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