3x^2-4x=68

Solution for 3x^2-4x=68 equation:

3x^2-4x=68

We move all terms to the left:

3x^2-4x-(68)=0

a = 3; b = -4; c = -68;
Δ = b2-4ac
Δ = -42-4·3·(-68)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{13}}{2*3}=\frac{4-8\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{13}}{2*3}=\frac{4+8\sqrt{13}}{6}
$