2y^+11y=y^2+3y-28

Solution for 2y^+11y=y^2+3y-28 equation:

2y^+11y=y^2+3y-28

We move all terms to the left:

2y^+11y-(y^2+3y-28)=0

We add all the numbers together, and all the variables

13y-(y^2+3y-28)=0

We get rid of parentheses

-y^2+13y-3y+28=0

We add all the numbers together, and all the variables

-1y^2+10y+28=0

a = -1; b = 10; c = +28;
Δ = b2-4ac
Δ = 102-4·(-1)·28
Δ = 212
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:

$\sqrt{\Delta}=\sqrt{212}=\sqrt{4*53}=\sqrt{4}*\sqrt{53}=2\sqrt{53}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{53}}{2*-1}=\frac{-10-2\sqrt{53}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{53}}{2*-1}=\frac{-10+2\sqrt{53}}{-2}
$