(3x+1)(x+3)=2x2+10

Solution for (3x+1)(x+3)=2x2+10 equation:

(3x+1)(x+3)=2x^2+10

We move all terms to the left:

(3x+1)(x+3)-(2x^2+10)=0

We get rid of parentheses

-2x^2+(3x+1)(x+3)-10=0

We multiply parentheses ..

-2x^2+(+3x^2+9x+x+3)-10=0

We get rid of parentheses

-2x^2+3x^2+9x+x+3-10=0

We add all the numbers together, and all the variables

x^2+10x-7=0

a = 1; b = 10; c = -7;
Δ = b2-4ac
Δ = 102-4·1·(-7)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8\sqrt{2}}{2*1}=\frac{-10-8\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8\sqrt{2}}{2*1}=\frac{-10+8\sqrt{2}}{2}
$