5y^2+24y+27=0

Solution for 5y^2+24y+27=0 equation:

5y^2+24y+27=0

a = 5; b = 24; c = +27;
Δ = b2-4ac
Δ = 242-4·5·27
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6}{2*5}=\frac{-30}{10}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6}{2*5}=\frac{-18}{10}
=-1+4/5
$