4x-13x^2+36=0

Solution for 4x-13x^2+36=0 equation:

4x-13x^2+36=0

a = -13; b = 4; c = +36;
Δ = b2-4ac
Δ = 42-4·(-13)·36
Δ = 1888
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:

$\sqrt{\Delta}=\sqrt{1888}=\sqrt{16*118}=\sqrt{16}*\sqrt{118}=4\sqrt{118}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{118}}{2*-13}=\frac{-4-4\sqrt{118}}{-26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{118}}{2*-13}=\frac{-4+4\sqrt{118}}{-26}
$