0=x^2+28x-5

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Solution for 0=x^2+28x-5 equation:



0=x^2+28x-5
We move all terms to the left:
0-(x^2+28x-5)=0
We add all the numbers together, and all the variables
-(x^2+28x-5)=0
We get rid of parentheses
-x^2-28x+5=0
We add all the numbers together, and all the variables
-1x^2-28x+5=0
a = -1; b = -28; c = +5;
Δ = b2-4ac
Δ = -282-4·(-1)·5
Δ = 804
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:
$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2\sqrt{201}}{2*-1}=\frac{28-2\sqrt{201}}{-2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2\sqrt{201}}{2*-1}=\frac{28+2\sqrt{201}}{-2} $

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