8+(-6x)+(1x^2)=0

Solution for 8+(-6x)+(1x^2)=0 equation:

8+(-6x)+(1x^2)=0

determiningTheFunctionDomain
1x^2+(-6x)+8=0

We add all the numbers together, and all the variables

x^2+(-6x)+8=0

We get rid of parentheses

x^2-6x+8=0

a = 1; b = -6; c = +8;
Δ = b2-4ac
Δ = -62-4·1·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2}{2*1}=\frac{8}{2}
=4
$