(12x^2-18x-20)/((4x+3)^2)=0

Solution for (12x^2-18x-20)/((4x+3)^2)=0 equation:

(12x^2-18x-20)/((4x+3)^2)=0


Domain of the equation: ((4x+3)^2)!=0

x∈R


We multiply all the terms by the denominator


(12x^2-18x-20)=0

We get rid of parentheses

12x^2-18x-20=0

a = 12; b = -18; c = -20;
Δ = b2-4ac
Δ = -182-4·12·(-20)
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{321}}{2*12}=\frac{18-2\sqrt{321}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{321}}{2*12}=\frac{18+2\sqrt{321}}{24}
$