5(2x+10)=x^2

Solution for 5(2x+10)=x^2 equation:

5(2x+10)=x^2

We move all terms to the left:

5(2x+10)-(x^2)=0

determiningTheFunctionDomain
-x^2+5(2x+10)=0

We add all the numbers together, and all the variables

-1x^2+5(2x+10)=0

We multiply parentheses

-1x^2+10x+50=0

a = -1; b = 10; c = +50;
Δ = b2-4ac
Δ = 102-4·(-1)·50
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{3}}{2*-1}=\frac{-10-10\sqrt{3}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{3}}{2*-1}=\frac{-10+10\sqrt{3}}{-2}
$