4x^2+x-23=0

Solution for 4x^2+x-23=0 equation:

4x^2+x-23=0

a = 4; b = 1; c = -23;
Δ = b2-4ac
Δ = 12-4·4·(-23)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{41}}{2*4}=\frac{-1-3\sqrt{41}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{41}}{2*4}=\frac{-1+3\sqrt{41}}{8}
$