4y^2+22y+18=0

Solution for 4y^2+22y+18=0 equation:

4y^2+22y+18=0

a = 4; b = 22; c = +18;
Δ = b2-4ac
Δ = 222-4·4·18
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*4}=\frac{-36}{8}
=-4+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*4}=\frac{-8}{8}
=-1
$