(2y-16)(2y-18)=150

Solution for (2y-16)(2y-18)=150 equation:

(2y-16)(2y-18)=150

We move all terms to the left:

(2y-16)(2y-18)-(150)=0

We multiply parentheses ..

(+4y^2-36y-32y+288)-150=0

We get rid of parentheses

4y^2-36y-32y+288-150=0

We add all the numbers together, and all the variables

4y^2-68y+138=0

a = 4; b = -68; c = +138;
Δ = b2-4ac
Δ = -682-4·4·138
Δ = 2416
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2416}=\sqrt{16*151}=\sqrt{16}*\sqrt{151}=4\sqrt{151}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-68)-4\sqrt{151}}{2*4}=\frac{68-4\sqrt{151}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+4\sqrt{151}}{2*4}=\frac{68+4\sqrt{151}}{8}
$