5(x+4)=x^2+6

Solution for 5(x+4)=x^2+6 equation:

5(x+4)=x^2+6

We move all terms to the left:

5(x+4)-(x^2+6)=0

We multiply parentheses

5x-(x^2+6)+20=0

We get rid of parentheses

-x^2+5x-6+20=0

We add all the numbers together, and all the variables

-1x^2+5x+14=0

a = -1; b = 5; c = +14;
Δ = b2-4ac
Δ = 52-4·(-1)·14
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-9}{2*-1}=\frac{-14}{-2}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+9}{2*-1}=\frac{4}{-2}
=-2
$