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2y^2+10y+12=0
a = 2; b = 10; c = +12;
Δ = b2-4ac
Δ = 102-4·2·12
Δ = 4
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{4}=2$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*2}=\frac{-12}{4} =-3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*2}=\frac{-8}{4} =-2 $
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