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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


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

(6y^2-5)

We get rid of parentheses

6y^2-5

Back to the equation:

-(6y^2-5)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-2y^2-5y+7=0

a = -2; b = -5; c = +7;
Δ = b2-4ac
Δ = -52-4·(-2)·7
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*-2}=\frac{-4}{-4}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*-2}=\frac{14}{-4}
=-3+1/2
$