24x^2-40x+15=0

Solution for 24x^2-40x+15=0 equation:

24x^2-40x+15=0

a = 24; b = -40; c = +15;
Δ = b2-4ac
Δ = -402-4·24·15
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{10}}{2*24}=\frac{40-4\sqrt{10}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{10}}{2*24}=\frac{40+4\sqrt{10}}{48}
$