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3=2y^2+5y
We move all terms to the left:
3-(2y^2+5y)=0
We get rid of parentheses
-2y^2-5y+3=0
a = -2; b = -5; c = +3;
Δ = b2-4ac
Δ = -52-4·(-2)·3
Δ = 49
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{49}=7$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*-2}=\frac{-2}{-4} =1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*-2}=\frac{12}{-4} =-3 $
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