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5y^2-3y=y
We move all terms to the left:
5y^2-3y-(y)=0
We add all the numbers together, and all the variables
5y^2-4y=0
a = 5; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·5·0
Δ = 16
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*5}=\frac{0}{10} =0 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*5}=\frac{8}{10} =4/5 $
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