(3x-2)(4x-5)=5x^2+12

Solution for (3x-2)(4x-5)=5x^2+12 equation:

(3x-2)(4x-5)=5x^2+12

We move all terms to the left:

(3x-2)(4x-5)-(5x^2+12)=0

We get rid of parentheses

-5x^2+(3x-2)(4x-5)-12=0

We multiply parentheses ..

-5x^2+(+12x^2-15x-8x+10)-12=0

We get rid of parentheses

-5x^2+12x^2-15x-8x+10-12=0

We add all the numbers together, and all the variables

7x^2-23x-2=0

a = 7; b = -23; c = -2;
Δ = b2-4ac
Δ = -232-4·7·(-2)
Δ = 585
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:

$\sqrt{\Delta}=\sqrt{585}=\sqrt{9*65}=\sqrt{9}*\sqrt{65}=3\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-3\sqrt{65}}{2*7}=\frac{23-3\sqrt{65}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+3\sqrt{65}}{2*7}=\frac{23+3\sqrt{65}}{14}
$