(3)(16x^2+24x-43)=0

Solution for (3)(16x^2+24x-43)=0 equation:

(3)(16x^2+24x-43)=0

We multiply parentheses

48x^2+72x-129=0

a = 48; b = 72; c = -129;
Δ = b2-4ac
Δ = 722-4·48·(-129)
Δ = 29952
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:

$\sqrt{\Delta}=\sqrt{29952}=\sqrt{2304*13}=\sqrt{2304}*\sqrt{13}=48\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-48\sqrt{13}}{2*48}=\frac{-72-48\sqrt{13}}{96}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+48\sqrt{13}}{2*48}=\frac{-72+48\sqrt{13}}{96}
$