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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3x^2+6x-9=0

a = 3; b = 6; c = -9;
Δ = b2-4ac
Δ = 62-4·3·(-9)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12}{2*3}=\frac{-18}{6}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12}{2*3}=\frac{6}{6}
=1
$