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4y^2+2y-40=0
a = 4; b = 2; c = -40;
Δ = b2-4ac
Δ = 22-4·4·(-40)
Δ = 644
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{161}}{2*4}=\frac{-2-2\sqrt{161}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{161}}{2*4}=\frac{-2+2\sqrt{161}}{8} $
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