(3x^2-4x=5)=2x3=6x-2)

Solution for (3x^2-4x=5)=2x3=6x-2) equation:

(3x^2-4x=5)=2x^3=6x-2)

We move all terms to the left:

(3x^2-4x-(5))=0

We get rid of parentheses

3x^2-4x-5=0

a = 3; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·3·(-5)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{19}}{2*3}=\frac{4-2\sqrt{19}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{19}}{2*3}=\frac{4+2\sqrt{19}}{6}
$