(3y^2-6y+5)=(-4y^2+7y+2)

Solution for (3y^2-6y+5)=(-4y^2+7y+2) equation:

(3y^2-6y+5)=(-4y^2+7y+2)

We move all terms to the left:

(3y^2-6y+5)-((-4y^2+7y+2))=0

We get rid of parentheses

-((-4y^2+7y+2))+3y^2-6y+5=0


We calculate terms in parentheses: -((-4y^2+7y+2)), so:

(-4y^2+7y+2)

We get rid of parentheses

-4y^2+7y+2

Back to the equation:

-(-4y^2+7y+2)


We add all the numbers together, and all the variables

3y^2-(-4y^2+7y+2)-6y+5=0

We get rid of parentheses

3y^2+4y^2-7y-6y-2+5=0

We add all the numbers together, and all the variables

7y^2-13y+3=0

a = 7; b = -13; c = +3;
Δ = b2-4ac
Δ = -132-4·7·3
Δ = 85
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{85}}{2*7}=\frac{13-\sqrt{85}}{14}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{85}}{2*7}=\frac{13+\sqrt{85}}{14}
$