(-3y^2+y-2)+(5y^2+6y+7=)

Solution for (-3y^2+y-2)+(5y^2+6y+7=) equation:

(-3y^2+y-2)+(5y^2+6y+7=)

We move all terms to the left:

(-3y^2+y-2)+(5y^2+6y+7-())=0

We get rid of parentheses

-3y^2+y+(5y^2+6y+7-())-2=0


We calculate terms in parentheses: +(5y^2+6y+7-()), so:

5y^2+6y+7-()

We add all the numbers together, and all the variables

5y^2+6y

Back to the equation:

+(5y^2+6y)


We get rid of parentheses

-3y^2+5y^2+y+6y-2=0

We add all the numbers together, and all the variables

2y^2+7y-2=0

a = 2; b = 7; c = -2;
Δ = b2-4ac
Δ = 72-4·2·(-2)
Δ = 65
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{-(7)-\sqrt{65}}{2*2}=\frac{-7-\sqrt{65}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{65}}{2*2}=\frac{-7+\sqrt{65}}{4}
$