7x^2+32x-576=0

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Solution for 7x^2+32x-576=0 equation:



7x^2+32x-576=0
a = 7; b = 32; c = -576;
Δ = b2-4ac
Δ = 322-4·7·(-576)
Δ = 17152
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:
$\sqrt{\Delta}=\sqrt{17152}=\sqrt{256*67}=\sqrt{256}*\sqrt{67}=16\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{67}}{2*7}=\frac{-32-16\sqrt{67}}{14} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{67}}{2*7}=\frac{-32+16\sqrt{67}}{14} $

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