8y+3y(y-5)=3(y+2)-2

Solution for 8y+3y(y-5)=3(y+2)-2 equation:

8y+3y(y-5)=3(y+2)-2

We move all terms to the left:

8y+3y(y-5)-(3(y+2)-2)=0

We multiply parentheses

3y^2+8y-15y-(3(y+2)-2)=0


We calculate terms in parentheses: -(3(y+2)-2), so:

3(y+2)-2

We multiply parentheses

3y+6-2

We add all the numbers together, and all the variables

3y+4

Back to the equation:

-(3y+4)


We add all the numbers together, and all the variables

3y^2-7y-(3y+4)=0

We get rid of parentheses

3y^2-7y-3y-4=0

We add all the numbers together, and all the variables

3y^2-10y-4=0

a = 3; b = -10; c = -4;
Δ = b2-4ac
Δ = -102-4·3·(-4)
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{37}}{2*3}=\frac{10-2\sqrt{37}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{37}}{2*3}=\frac{10+2\sqrt{37}}{6}
$