24x2+42x-15=0

Solution for 24x2+42x-15=0 equation:

24x^2+42x-15=0

a = 24; b = 42; c = -15;
Δ = b2-4ac
Δ = 422-4·24·(-15)
Δ = 3204
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:

$\sqrt{\Delta}=\sqrt{3204}=\sqrt{36*89}=\sqrt{36}*\sqrt{89}=6\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{89}}{2*24}=\frac{-42-6\sqrt{89}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{89}}{2*24}=\frac{-42+6\sqrt{89}}{48}
$