12/5+y^2=25

Solution for 12/5+y^2=25 equation:

12/5+y^2=25

We move all terms to the left:

12/5+y^2-(25)=0

determiningTheFunctionDomain
y^2-25+12/5=0

We multiply all the terms by the denominator


y^2*5+12-25*5=0

We add all the numbers together, and all the variables

y^2*5-113=0

Wy multiply elements

5y^2-113=0

a = 5; b = 0; c = -113;
Δ = b2-4ac
Δ = 02-4·5·(-113)
Δ = 2260
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:

$\sqrt{\Delta}=\sqrt{2260}=\sqrt{4*565}=\sqrt{4}*\sqrt{565}=2\sqrt{565}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{565}}{2*5}=\frac{0-2\sqrt{565}}{10}
=-\frac{2\sqrt{565}}{10}
=-\frac{\sqrt{565}}{5}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{565}}{2*5}=\frac{0+2\sqrt{565}}{10}
=\frac{2\sqrt{565}}{10}
=\frac{\sqrt{565}}{5}
$