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2y^2+8y=-5
We move all terms to the left:
2y^2+8y-(-5)=0
We add all the numbers together, and all the variables
2y^2+8y+5=0
a = 2; b = 8; c = +5;
Δ = b2-4ac
Δ = 82-4·2·5
Δ = 24
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{6}}{2*2}=\frac{-8-2\sqrt{6}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{6}}{2*2}=\frac{-8+2\sqrt{6}}{4} $
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