4x^2+20x-234=0

Solution for 4x^2+20x-234=0 equation:

4x^2+20x-234=0

a = 4; b = 20; c = -234;
Δ = b2-4ac
Δ = 202-4·4·(-234)
Δ = 4144
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:

$\sqrt{\Delta}=\sqrt{4144}=\sqrt{16*259}=\sqrt{16}*\sqrt{259}=4\sqrt{259}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{259}}{2*4}=\frac{-20-4\sqrt{259}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{259}}{2*4}=\frac{-20+4\sqrt{259}}{8}
$