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

Solution for 2(2^2)-5y(+2)=0 equation:

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

We add all the numbers together, and all the variables

-5y2+22^2=0

We add all the numbers together, and all the variables

-5y^2+484=0

a = -5; b = 0; c = +484;
Δ = b2-4ac
Δ = 02-4·(-5)·484
Δ = 9680
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:

$\sqrt{\Delta}=\sqrt{9680}=\sqrt{1936*5}=\sqrt{1936}*\sqrt{5}=44\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-44\sqrt{5}}{2*-5}=\frac{0-44\sqrt{5}}{-10}
=-\frac{44\sqrt{5}}{-10}
=-\frac{22\sqrt{5}}{-5}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+44\sqrt{5}}{2*-5}=\frac{0+44\sqrt{5}}{-10}
=\frac{44\sqrt{5}}{-10}
=\frac{22\sqrt{5}}{-5}
$