2y(y+5)=7y+35

Solution for 2y(y+5)=7y+35 equation:

2y(y+5)=7y+35

We move all terms to the left:

2y(y+5)-(7y+35)=0

We multiply parentheses

2y^2+10y-(7y+35)=0

We get rid of parentheses

2y^2+10y-7y-35=0

We add all the numbers together, and all the variables

2y^2+3y-35=0

a = 2; b = 3; c = -35;
Δ = b2-4ac
Δ = 32-4·2·(-35)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-17}{2*2}=\frac{-20}{4}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+17}{2*2}=\frac{14}{4}
=3+1/2
$