3x+43=x^2+3

Solution for 3x+43=x^2+3 equation:

3x+43=x^2+3

We move all terms to the left:

3x+43-(x^2+3)=0

We get rid of parentheses

-x^2+3x-3+43=0

We add all the numbers together, and all the variables

-1x^2+3x+40=0

a = -1; b = 3; c = +40;
Δ = b2-4ac
Δ = 32-4·(-1)·40
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*-1}=\frac{-16}{-2}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*-1}=\frac{10}{-2}
=-5
$