5(y-4)(2y+16)=168

Solution for 5(y-4)(2y+16)=168 equation:

5(y-4)(2y+16)=168

We move all terms to the left:

5(y-4)(2y+16)-(168)=0

We multiply parentheses ..

5(+2y^2+16y-8y-64)-168=0

We multiply parentheses

10y^2+80y-40y-320-168=0

We add all the numbers together, and all the variables

10y^2+40y-488=0

a = 10; b = 40; c = -488;
Δ = b2-4ac
Δ = 402-4·10·(-488)
Δ = 21120
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:

$\sqrt{\Delta}=\sqrt{21120}=\sqrt{64*330}=\sqrt{64}*\sqrt{330}=8\sqrt{330}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{330}}{2*10}=\frac{-40-8\sqrt{330}}{20}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{330}}{2*10}=\frac{-40+8\sqrt{330}}{20}
$