3x^2+40=121

Solution for 3x^2+40=121 equation:

3x^2+40=121

We move all terms to the left:

3x^2+40-(121)=0

We add all the numbers together, and all the variables

3x^2-81=0

a = 3; b = 0; c = -81;
Δ = b2-4ac
Δ = 02-4·3·(-81)
Δ = 972
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:

$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{3}}{2*3}=\frac{0-18\sqrt{3}}{6}
=-\frac{18\sqrt{3}}{6}
=-3\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{3}}{2*3}=\frac{0+18\sqrt{3}}{6}
=\frac{18\sqrt{3}}{6}
=3\sqrt{3}
$