3x^2=6x+23

Solution for 3x^2=6x+23 equation:

3x^2=6x+23

We move all terms to the left:

3x^2-(6x+23)=0

We get rid of parentheses

3x^2-6x-23=0

a = 3; b = -6; c = -23;
Δ = b2-4ac
Δ = -62-4·3·(-23)
Δ = 312
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:

$\sqrt{\Delta}=\sqrt{312}=\sqrt{4*78}=\sqrt{4}*\sqrt{78}=2\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{78}}{2*3}=\frac{6-2\sqrt{78}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{78}}{2*3}=\frac{6+2\sqrt{78}}{6}
$