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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

3y(y+5)-(-1y+7)=0

We multiply parentheses

3y^2+15y-(-1y+7)=0

We get rid of parentheses

3y^2+15y+1y-7=0

We add all the numbers together, and all the variables

3y^2+16y-7=0

a = 3; b = 16; c = -7;
Δ = b2-4ac
Δ = 162-4·3·(-7)
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{85}}{2*3}=\frac{-16-2\sqrt{85}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{85}}{2*3}=\frac{-16+2\sqrt{85}}{6}
$