(7y^2-y-6)=-(4y^2-3y+2)

Solution for (7y^2-y-6)=-(4y^2-3y+2) equation:

(7y^2-y-6)=-(4y^2-3y+2)

We move all terms to the left:

(7y^2-y-6)-(-(4y^2-3y+2))=0

We get rid of parentheses

7y^2-y-(-(4y^2-3y+2))-6=0


We calculate terms in parentheses: -(-(4y^2-3y+2)), so:

-(4y^2-3y+2)

We get rid of parentheses

-4y^2+3y-2

Back to the equation:

-(-4y^2+3y-2)


We add all the numbers together, and all the variables

7y^2-(-4y^2+3y-2)-1y-6=0

We get rid of parentheses

7y^2+4y^2-3y-1y+2-6=0

We add all the numbers together, and all the variables

11y^2-4y-4=0

a = 11; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·11·(-4)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{3}}{2*11}=\frac{4-8\sqrt{3}}{22}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{3}}{2*11}=\frac{4+8\sqrt{3}}{22}
$