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(x-6)(x+2)=0.
We multiply parentheses ..
(+x^2+2x-6x-12)=0
We get rid of parentheses
x^2+2x-6x-12=0
We add all the numbers together, and all the variables
x^2-4x-12=0
a = 1; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·1·(-12)
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*1}=\frac{-4}{2} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*1}=\frac{12}{2} =6 $
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