78=(2x+3)(3x-4)

Solution for 78=(2x+3)(3x-4) equation:

78=(2x+3)(3x-4)

We move all terms to the left:

78-((2x+3)(3x-4))=0

We multiply parentheses ..

-((+6x^2-8x+9x-12))+78=0


We calculate terms in parentheses: -((+6x^2-8x+9x-12)), so:

(+6x^2-8x+9x-12)

We get rid of parentheses

6x^2-8x+9x-12

We add all the numbers together, and all the variables

6x^2+x-12

Back to the equation:

-(6x^2+x-12)


We get rid of parentheses

-6x^2-x+12+78=0

We add all the numbers together, and all the variables

-6x^2-1x+90=0

a = -6; b = -1; c = +90;
Δ = b2-4ac
Δ = -12-4·(-6)·90
Δ = 2161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{2161}}{2*-6}=\frac{1-\sqrt{2161}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{2161}}{2*-6}=\frac{1+\sqrt{2161}}{-12}
$