5y^2+16-2y=(4y^2-2y+24)

Solution for 5y^2+16-2y=(4y^2-2y+24) equation:

5y^2+16-2y=(4y^2-2y+24)

We move all terms to the left:

5y^2+16-2y-((4y^2-2y+24))=0


We calculate terms in parentheses: -((4y^2-2y+24)), so:

(4y^2-2y+24)

We get rid of parentheses

4y^2-2y+24

Back to the equation:

-(4y^2-2y+24)


We get rid of parentheses

5y^2-4y^2-2y+2y-24+16=0

We add all the numbers together, and all the variables

y^2-8=0

a = 1; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·1·(-8)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*1}=\frac{0-4\sqrt{2}}{2}
=-\frac{4\sqrt{2}}{2}
=-2\sqrt{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*1}=\frac{0+4\sqrt{2}}{2}
=\frac{4\sqrt{2}}{2}
=2\sqrt{2}
$