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4x^2+24x+16=0
a = 4; b = 24; c = +16;
Δ = b2-4ac
Δ = 242-4·4·16
Δ = 320
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{5}}{2*4}=\frac{-24-8\sqrt{5}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{5}}{2*4}=\frac{-24+8\sqrt{5}}{8} $
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