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2y^2+y2+5y-16=52
We move all terms to the left:
2y^2+y2+5y-16-(52)=0
We add all the numbers together, and all the variables
3y^2+5y-68=0
a = 3; b = 5; c = -68;
Δ = b2-4ac
Δ = 52-4·3·(-68)
Δ = 841
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{841}=29$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-29}{2*3}=\frac{-34}{6} =-5+2/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+29}{2*3}=\frac{24}{6} =4 $
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