(4y+10)(y-3y)=5

Solution for (4y+10)(y-3y)=5 equation:

(4y+10)(y-3y)=5

We move all terms to the left:

(4y+10)(y-3y)-(5)=0

We add all the numbers together, and all the variables

(4y+10)(-2y)-5=0

We multiply parentheses ..

(-8y^2-20y)-5=0

We get rid of parentheses

-8y^2-20y-5=0

a = -8; b = -20; c = -5;
Δ = b2-4ac
Δ = -202-4·(-8)·(-5)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{15}}{2*-8}=\frac{20-4\sqrt{15}}{-16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{15}}{2*-8}=\frac{20+4\sqrt{15}}{-16}
$