7y^2+22y+3=0

Solution for 7y^2+22y+3=0 equation:

7y^2+22y+3=0

a = 7; b = 22; c = +3;
Δ = b2-4ac
Δ = 222-4·7·3
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-20}{2*7}=\frac{-42}{14}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+20}{2*7}=\frac{-2}{14}
=-1/7
$