24x^2+16x-3=0

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

24x^2+16x-3=0

a = 24; b = 16; c = -3;
Δ = b2-4ac
Δ = 162-4·24·(-3)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{34}}{2*24}=\frac{-16-4\sqrt{34}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{34}}{2*24}=\frac{-16+4\sqrt{34}}{48}
$