3x+38=x^2+10

Solution for 3x+38=x^2+10 equation:

3x+38=x^2+10

We move all terms to the left:

3x+38-(x^2+10)=0

We get rid of parentheses

-x^2+3x-10+38=0

We add all the numbers together, and all the variables

-1x^2+3x+28=0

a = -1; b = 3; c = +28;
Δ = b2-4ac
Δ = 32-4·(-1)·28
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-11}{2*-1}=\frac{-14}{-2}
=+7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*-1}=\frac{8}{-2}
=-4
$