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2y^2+25y+78=0
a = 2; b = 25; c = +78;
Δ = b2-4ac
Δ = 252-4·2·78
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-1}{2*2}=\frac{-26}{4} =-6+1/2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+1}{2*2}=\frac{-24}{4} =-6 $
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