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2y^2+3y-5=4
We move all terms to the left:
2y^2+3y-5-(4)=0
We add all the numbers together, and all the variables
2y^2+3y-9=0
a = 2; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·2·(-9)
Δ = 81
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{81}=9$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*2}=\frac{-12}{4} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*2}=\frac{6}{4} =1+1/2 $
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