(3x+4)(4x-10)=9

Solution for (3x+4)(4x-10)=9 equation:

(3x+4)(4x-10)=9

We move all terms to the left:

(3x+4)(4x-10)-(9)=0

We multiply parentheses ..

(+12x^2-30x+16x-40)-9=0

We get rid of parentheses

12x^2-30x+16x-40-9=0

We add all the numbers together, and all the variables

12x^2-14x-49=0

a = 12; b = -14; c = -49;
Δ = b2-4ac
Δ = -142-4·12·(-49)
Δ = 2548
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:

$\sqrt{\Delta}=\sqrt{2548}=\sqrt{196*13}=\sqrt{196}*\sqrt{13}=14\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14\sqrt{13}}{2*12}=\frac{14-14\sqrt{13}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14\sqrt{13}}{2*12}=\frac{14+14\sqrt{13}}{24}
$