21x^2+43x+14=0

Solution for 21x^2+43x+14=0 equation:

21x^2+43x+14=0

a = 21; b = 43; c = +14;
Δ = b2-4ac
Δ = 432-4·21·14
Δ = 673
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-\sqrt{673}}{2*21}=\frac{-43-\sqrt{673}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+\sqrt{673}}{2*21}=\frac{-43+\sqrt{673}}{42}
$