x^2+14x=128

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Solution for x^2+14x=128 equation:



x^2+14x=128
We move all terms to the left:
x^2+14x-(128)=0
a = 1; b = 14; c = -128;
Δ = b2-4ac
Δ = 142-4·1·(-128)
Δ = 708
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:
$\sqrt{\Delta}=\sqrt{708}=\sqrt{4*177}=\sqrt{4}*\sqrt{177}=2\sqrt{177}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{177}}{2*1}=\frac{-14-2\sqrt{177}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{177}}{2*1}=\frac{-14+2\sqrt{177}}{2} $

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