0=3x^2+14x-23

Solution for 0=3x^2+14x-23 equation:

0=3x^2+14x-23

We move all terms to the left:

0-(3x^2+14x-23)=0

We add all the numbers together, and all the variables

-(3x^2+14x-23)=0

We get rid of parentheses

-3x^2-14x+23=0

a = -3; b = -14; c = +23;
Δ = b2-4ac
Δ = -142-4·(-3)·23
Δ = 472
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:

$\sqrt{\Delta}=\sqrt{472}=\sqrt{4*118}=\sqrt{4}*\sqrt{118}=2\sqrt{118}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{118}}{2*-3}=\frac{14-2\sqrt{118}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{118}}{2*-3}=\frac{14+2\sqrt{118}}{-6}
$