78=(2x+5)(3x-4)

Solution for 78=(2x+5)(3x-4) equation:

78=(2x+5)(3x-4)

We move all terms to the left:

78-((2x+5)(3x-4))=0

We multiply parentheses ..

-((+6x^2-8x+15x-20))+78=0


We calculate terms in parentheses: -((+6x^2-8x+15x-20)), so:

(+6x^2-8x+15x-20)

We get rid of parentheses

6x^2-8x+15x-20

We add all the numbers together, and all the variables

6x^2+7x-20

Back to the equation:

-(6x^2+7x-20)


We get rid of parentheses

-6x^2-7x+20+78=0

We add all the numbers together, and all the variables

-6x^2-7x+98=0

a = -6; b = -7; c = +98;
Δ = b2-4ac
Δ = -72-4·(-6)·98
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-49}{2*-6}=\frac{-42}{-12}
=3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+49}{2*-6}=\frac{56}{-12}
=-4+2/3
$