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15y^2+16y+4=0
a = 15; b = 16; c = +4;
Δ = b2-4ac
Δ = 162-4·15·4
Δ = 16
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{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*15}=\frac{-20}{30} =-2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*15}=\frac{-12}{30} =-2/5 $
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