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2y^2+12y-7=0
a = 2; b = 12; c = -7;
Δ = b2-4ac
Δ = 122-4·2·(-7)
Δ = 200
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-10\sqrt{2}}{2*2}=\frac{-12-10\sqrt{2}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+10\sqrt{2}}{2*2}=\frac{-12+10\sqrt{2}}{4} $
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