24x^2+42x+10=0

Solution for 24x^2+42x+10=0 equation:

24x^2+42x+10=0

a = 24; b = 42; c = +10;
Δ = b2-4ac
Δ = 422-4·24·10
Δ = 804
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:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{201}}{2*24}=\frac{-42-2\sqrt{201}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{201}}{2*24}=\frac{-42+2\sqrt{201}}{48}
$