21x^2+14x-9=0

Solution for 21x^2+14x-9=0 equation:

21x^2+14x-9=0

a = 21; b = 14; c = -9;
Δ = b2-4ac
Δ = 142-4·21·(-9)
Δ = 952
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:

$\sqrt{\Delta}=\sqrt{952}=\sqrt{4*238}=\sqrt{4}*\sqrt{238}=2\sqrt{238}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{238}}{2*21}=\frac{-14-2\sqrt{238}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{238}}{2*21}=\frac{-14+2\sqrt{238}}{42}
$