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4y^2+68y-289=0
a = 4; b = 68; c = -289;
Δ = b2-4ac
Δ = 682-4·4·(-289)
Δ = 9248
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9248}=\sqrt{4624*2}=\sqrt{4624}*\sqrt{2}=68\sqrt{2}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-68\sqrt{2}}{2*4}=\frac{-68-68\sqrt{2}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+68\sqrt{2}}{2*4}=\frac{-68+68\sqrt{2}}{8} $
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