3-((5y^2)/4)=0,y

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Solution for 3-((5y^2)/4)=0,y equation:



3-((5y^2)/4)=0.y
We move all terms to the left:
3-((5y^2)/4)-(0.y)=0
We add all the numbers together, and all the variables
-(5y^2/4)+3-0=0
We add all the numbers together, and all the variables
-(5y^2/4)+3=0
We get rid of parentheses
-5y^2/4+3=0
We multiply all the terms by the denominator
-5y^2+3*4=0
We add all the numbers together, and all the variables
-5y^2+12=0
a = -5; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-5)·12
Δ = 240
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:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*-5}=\frac{0-4\sqrt{15}}{-10} =-\frac{4\sqrt{15}}{-10} =-\frac{2\sqrt{15}}{-5} $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*-5}=\frac{0+4\sqrt{15}}{-10} =\frac{4\sqrt{15}}{-10} =\frac{2\sqrt{15}}{-5} $

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