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3/5y+y=48
We move all terms to the left:
3/5y+y-(48)=0
Domain of the equation: 5y!=0We add all the numbers together, and all the variables
y!=0/5
y!=0
y∈R
y+3/5y-48=0
We multiply all the terms by the denominator
y*5y-48*5y+3=0
Wy multiply elements
5y^2-240y+3=0
a = 5; b = -240; c = +3;
Δ = b2-4ac
Δ = -2402-4·5·3
Δ = 57540
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{57540}=\sqrt{4*14385}=\sqrt{4}*\sqrt{14385}=2\sqrt{14385}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-2\sqrt{14385}}{2*5}=\frac{240-2\sqrt{14385}}{10} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+2\sqrt{14385}}{2*5}=\frac{240+2\sqrt{14385}}{10} $
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