2(3x+5)=x^2

Solution for 2(3x+5)=x^2 equation:

2(3x+5)=x^2

We move all terms to the left:

2(3x+5)-(x^2)=0

determiningTheFunctionDomain
-x^2+2(3x+5)=0

We add all the numbers together, and all the variables

-1x^2+2(3x+5)=0

We multiply parentheses

-1x^2+6x+10=0

a = -1; b = 6; c = +10;
Δ = b2-4ac
Δ = 62-4·(-1)·10
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{19}}{2*-1}=\frac{-6-2\sqrt{19}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{19}}{2*-1}=\frac{-6+2\sqrt{19}}{-2}
$