5y^2+27y+34=0

Solution for 5y^2+27y+34=0 equation:

5y^2+27y+34=0

a = 5; b = 27; c = +34;
Δ = b2-4ac
Δ = 272-4·5·34
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-7}{2*5}=\frac{-34}{10}
=-3+2/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+7}{2*5}=\frac{-20}{10}
=-2
$