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4x^2+18x-80=0
a = 4; b = 18; c = -80;
Δ = b2-4ac
Δ = 182-4·4·(-80)
Δ = 1604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1604}=\sqrt{4*401}=\sqrt{4}*\sqrt{401}=2\sqrt{401}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{401}}{2*4}=\frac{-18-2\sqrt{401}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{401}}{2*4}=\frac{-18+2\sqrt{401}}{8} $
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