3x^2+13=40

Solution for 3x^2+13=40 equation:

3x^2+13=40

We move all terms to the left:

3x^2+13-(40)=0

We add all the numbers together, and all the variables

3x^2-27=0

a = 3; b = 0; c = -27;
Δ = b2-4ac
Δ = 02-4·3·(-27)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18}{2*3}=\frac{-18}{6}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18}{2*3}=\frac{18}{6}
=3
$