36x^2+72x-13=0

Solution for 36x^2+72x-13=0 equation:

36x^2+72x-13=0

a = 36; b = 72; c = -13;
Δ = b2-4ac
Δ = 722-4·36·(-13)
Δ = 7056
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7056}=84$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-84}{2*36}=\frac{-156}{72}
=-2+1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+84}{2*36}=\frac{12}{72}
=1/6
$