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(2/5)y-(3/5)y+5=-7
We move all terms to the left:
(2/5)y-(3/5)y+5-(-7)=0
Domain of the equation: 5)y!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
(+2/5)y-(+3/5)y+5-(-7)=0
We add all the numbers together, and all the variables
(+2/5)y-(+3/5)y+12=0
We multiply parentheses
2y^2-3y^2+12=0
We add all the numbers together, and all the variables
-1y^2+12=0
a = -1; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-1)·12
Δ = 48
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-1}=\frac{0-4\sqrt{3}}{-2} =-\frac{4\sqrt{3}}{-2} =-\frac{2\sqrt{3}}{-1} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-1}=\frac{0+4\sqrt{3}}{-2} =\frac{4\sqrt{3}}{-2} =\frac{2\sqrt{3}}{-1} $
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