(4y+11)(2y+19)=0

Solution for (4y+11)(2y+19)=0 equation:

(4y+11)(2y+19)=0

We multiply parentheses ..

(+8y^2+76y+22y+209)=0

We get rid of parentheses

8y^2+76y+22y+209=0

We add all the numbers together, and all the variables

8y^2+98y+209=0

a = 8; b = 98; c = +209;
Δ = b2-4ac
Δ = 982-4·8·209
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(98)-54}{2*8}=\frac{-152}{16}
=-9+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(98)+54}{2*8}=\frac{-44}{16}
=-2+3/4
$