5y^2=4-3y^2+5+y=

Solution for 5y^2=4-3y^2+5+y= equation:

5y^2=4-3y^2+5+y=

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

8y^2-1y-9=0

a = 8; b = -1; c = -9;
Δ = b2-4ac
Δ = -12-4·8·(-9)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*8}=\frac{-16}{16}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*8}=\frac{18}{16}
=1+1/8
$