5y^2+33y+40=0

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Solution for 5y^2+33y+40=0 equation:



5y^2+33y+40=0
a = 5; b = 33; c = +40;
Δ = b2-4ac
Δ = 332-4·5·40
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-17}{2*5}=\frac{-50}{10} =-5 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+17}{2*5}=\frac{-16}{10} =-1+3/5 $

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