(5x-4)(5x+4)+(6x)=100

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Solution for (5x-4)(5x+4)+(6x)=100 equation:



(5x-4)(5x+4)+(6x)=100
We move all terms to the left:
(5x-4)(5x+4)+(6x)-(100)=0
We add all the numbers together, and all the variables
6x+(5x-4)(5x+4)-100=0
We use the square of the difference formula
25x^2+6x-16-100=0
We add all the numbers together, and all the variables
25x^2+6x-116=0
a = 25; b = 6; c = -116;
Δ = b2-4ac
Δ = 62-4·25·(-116)
Δ = 11636
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:
$\sqrt{\Delta}=\sqrt{11636}=\sqrt{4*2909}=\sqrt{4}*\sqrt{2909}=2\sqrt{2909}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{2909}}{2*25}=\frac{-6-2\sqrt{2909}}{50} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{2909}}{2*25}=\frac{-6+2\sqrt{2909}}{50} $

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