7y+2y(y-5)=3(y+2)-5

Solution for 7y+2y(y-5)=3(y+2)-5 equation:

7y+2y(y-5)=3(y+2)-5

We move all terms to the left:

7y+2y(y-5)-(3(y+2)-5)=0

We multiply parentheses

2y^2+7y-10y-(3(y+2)-5)=0


We calculate terms in parentheses: -(3(y+2)-5), so:

3(y+2)-5

We multiply parentheses

3y+6-5

We add all the numbers together, and all the variables

3y+1

Back to the equation:

-(3y+1)


We add all the numbers together, and all the variables

2y^2-3y-(3y+1)=0

We get rid of parentheses

2y^2-3y-3y-1=0

We add all the numbers together, and all the variables

2y^2-6y-1=0

a = 2; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·2·(-1)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{11}}{2*2}=\frac{6-2\sqrt{11}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{11}}{2*2}=\frac{6+2\sqrt{11}}{4}
$