2/5y^2+3=11

Solution for 2/5y^2+3=11 equation:

2/5y^2+3=11

We move all terms to the left:

2/5y^2+3-(11)=0


Domain of the equation: 5y^2!=0

y^2!=0/5

y^2!=√0

y!=0

y∈R


We add all the numbers together, and all the variables

2/5y^2-8=0

We multiply all the terms by the denominator


-8*5y^2+2=0

Wy multiply elements

-40y^2+2=0

a = -40; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-40)·2
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*-40}=\frac{0-8\sqrt{5}}{-80}
=-\frac{8\sqrt{5}}{-80}
=-\frac{\sqrt{5}}{-10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*-40}=\frac{0+8\sqrt{5}}{-80}
=\frac{8\sqrt{5}}{-80}
=\frac{\sqrt{5}}{-10}
$