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3/5y+3=11+5y
We move all terms to the left:
3/5y+3-(11+5y)=0
Domain of the equation: 5y!=0We add all the numbers together, and all the variables
y!=0/5
y!=0
y∈R
3/5y-(5y+11)+3=0
We get rid of parentheses
3/5y-5y-11+3=0
We multiply all the terms by the denominator
-5y*5y-11*5y+3*5y+3=0
Wy multiply elements
-25y^2-55y+15y+3=0
We add all the numbers together, and all the variables
-25y^2-40y+3=0
a = -25; b = -40; c = +3;
Δ = b2-4ac
Δ = -402-4·(-25)·3
Δ = 1900
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1900}=\sqrt{100*19}=\sqrt{100}*\sqrt{19}=10\sqrt{19}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-10\sqrt{19}}{2*-25}=\frac{40-10\sqrt{19}}{-50} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+10\sqrt{19}}{2*-25}=\frac{40+10\sqrt{19}}{-50} $
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