y=(8y^2-5y+7)-(2y^2+7y+11)

Solution for y=(8y^2-5y+7)-(2y^2+7y+11) equation:

y=(8y^2-5y+7)-(2y^2+7y+11)

We move all terms to the left:

y-((8y^2-5y+7)-(2y^2+7y+11))=0


We calculate terms in parentheses: -((8y^2-5y+7)-(2y^2+7y+11)), so:

(8y^2-5y+7)-(2y^2+7y+11)

We get rid of parentheses

8y^2-2y^2-5y-7y+7-11

We add all the numbers together, and all the variables

6y^2-12y-4

Back to the equation:

-(6y^2-12y-4)


We get rid of parentheses

-6y^2+y+12y+4=0

We add all the numbers together, and all the variables

-6y^2+13y+4=0

a = -6; b = 13; c = +4;
Δ = b2-4ac
Δ = 132-4·(-6)·4
Δ = 265
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{265}}{2*-6}=\frac{-13-\sqrt{265}}{-12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{265}}{2*-6}=\frac{-13+\sqrt{265}}{-12}
$