13x^2+18x-5=0

Solution for 13x^2+18x-5=0 equation:

13x^2+18x-5=0

a = 13; b = 18; c = -5;
Δ = b2-4ac
Δ = 182-4·13·(-5)
Δ = 584
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:

$\sqrt{\Delta}=\sqrt{584}=\sqrt{4*146}=\sqrt{4}*\sqrt{146}=2\sqrt{146}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{146}}{2*13}=\frac{-18-2\sqrt{146}}{26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{146}}{2*13}=\frac{-18+2\sqrt{146}}{26}
$