11x^2+10x-5=0

Solution for 11x^2+10x-5=0 equation:

11x^2+10x-5=0

a = 11; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·11·(-5)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8\sqrt{5}}{2*11}=\frac{-10-8\sqrt{5}}{22}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8\sqrt{5}}{2*11}=\frac{-10+8\sqrt{5}}{22}
$