3x^2-4x=23

Solution for 3x^2-4x=23 equation:

3x^2-4x=23

We move all terms to the left:

3x^2-4x-(23)=0

a = 3; b = -4; c = -23;
Δ = b2-4ac
Δ = -42-4·3·(-23)
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{73}}{2*3}=\frac{4-2\sqrt{73}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{73}}{2*3}=\frac{4+2\sqrt{73}}{6}
$