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4y^2+6y-80=0
a = 4; b = 6; c = -80;
Δ = b2-4ac
Δ = 62-4·4·(-80)
Δ = 1316
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1316}=\sqrt{4*329}=\sqrt{4}*\sqrt{329}=2\sqrt{329}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{329}}{2*4}=\frac{-6-2\sqrt{329}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{329}}{2*4}=\frac{-6+2\sqrt{329}}{8} $
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