21x^2+47x-24=0

Solution for 21x^2+47x-24=0 equation:

21x^2+47x-24=0

a = 21; b = 47; c = -24;
Δ = b2-4ac
Δ = 472-4·21·(-24)
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4225}=65$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-65}{2*21}=\frac{-112}{42}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+65}{2*21}=\frac{18}{42}
=3/7
$