11x^2-142+13=0

Solution for 11x^2-142+13=0 equation:

11x^2-142+13=0

We add all the numbers together, and all the variables

11x^2-129=0

a = 11; b = 0; c = -129;
Δ = b2-4ac
Δ = 02-4·11·(-129)
Δ = 5676
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:

$\sqrt{\Delta}=\sqrt{5676}=\sqrt{4*1419}=\sqrt{4}*\sqrt{1419}=2\sqrt{1419}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1419}}{2*11}=\frac{0-2\sqrt{1419}}{22}
=-\frac{2\sqrt{1419}}{22}
=-\frac{\sqrt{1419}}{11}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1419}}{2*11}=\frac{0+2\sqrt{1419}}{22}
=\frac{2\sqrt{1419}}{22}
=\frac{\sqrt{1419}}{11}
$