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2y^2+160y=6300
We move all terms to the left:
2y^2+160y-(6300)=0
a = 2; b = 160; c = -6300;
Δ = b2-4ac
Δ = 1602-4·2·(-6300)
Δ = 76000
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{76000}=\sqrt{400*190}=\sqrt{400}*\sqrt{190}=20\sqrt{190}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-20\sqrt{190}}{2*2}=\frac{-160-20\sqrt{190}}{4} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+20\sqrt{190}}{2*2}=\frac{-160+20\sqrt{190}}{4} $
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