29x^2+20x-11=0

Solution for 29x^2+20x-11=0 equation:

29x^2+20x-11=0

a = 29; b = 20; c = -11;
Δ = b2-4ac
Δ = 202-4·29·(-11)
Δ = 1676
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:

$\sqrt{\Delta}=\sqrt{1676}=\sqrt{4*419}=\sqrt{4}*\sqrt{419}=2\sqrt{419}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{419}}{2*29}=\frac{-20-2\sqrt{419}}{58}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{419}}{2*29}=\frac{-20+2\sqrt{419}}{58}
$