4x^2+24x-256=0

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Solution for 4x^2+24x-256=0 equation:



4x^2+24x-256=0
a = 4; b = 24; c = -256;
Δ = b2-4ac
Δ = 242-4·4·(-256)
Δ = 4672
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:
$\sqrt{\Delta}=\sqrt{4672}=\sqrt{64*73}=\sqrt{64}*\sqrt{73}=8\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{73}}{2*4}=\frac{-24-8\sqrt{73}}{8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{73}}{2*4}=\frac{-24+8\sqrt{73}}{8} $

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