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4x^2+18x-16=0
a = 4; b = 18; c = -16;
Δ = b2-4ac
Δ = 182-4·4·(-16)
Δ = 580
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{145}}{2*4}=\frac{-18-2\sqrt{145}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{145}}{2*4}=\frac{-18+2\sqrt{145}}{8} $
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