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2y^2+35-19y=0.
a = 2; b = -19; c = +35;
Δ = b2-4ac
Δ = -192-4·2·35
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-9}{2*2}=\frac{10}{4} =2+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+9}{2*2}=\frac{28}{4} =7 $
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