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(x-4)(x+4)=6x
We move all terms to the left:
(x-4)(x+4)-(6x)=0
We add all the numbers together, and all the variables
-6x+(x-4)(x+4)=0
We use the square of the difference formula
x^2-6x-16=0
a = 1; b = -6; c = -16;
Δ = b2-4ac
Δ = -62-4·1·(-16)
Δ = 100
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*1}=\frac{-4}{2} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*1}=\frac{16}{2} =8 $
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