10-4y;2=2

Solution for 10-4y;2=2 equation:

10-4y^2=2

We move all terms to the left:

10-4y^2-(2)=0

We add all the numbers together, and all the variables

-4y^2+8=0

a = -4; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-4)·8
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*-4}=\frac{0-8\sqrt{2}}{-8}
=-\frac{8\sqrt{2}}{-8}
=-\frac{\sqrt{2}}{-1}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-4}=\frac{0+8\sqrt{2}}{-8}
=\frac{8\sqrt{2}}{-8}
=\frac{\sqrt{2}}{-1}
$