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4y^2+25+20y+y2-12y-25+3y-11=0
We add all the numbers together, and all the variables
5y^2+11y-11=0
a = 5; b = 11; c = -11;
Δ = b2-4ac
Δ = 112-4·5·(-11)
Δ = 341
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{341}}{2*5}=\frac{-11-\sqrt{341}}{10} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{341}}{2*5}=\frac{-11+\sqrt{341}}{10} $
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