21x^2+32x+7=0

Solution for 21x^2+32x+7=0 equation:

21x^2+32x+7=0

a = 21; b = 32; c = +7;
Δ = b2-4ac
Δ = 322-4·21·7
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{109}}{2*21}=\frac{-32-2\sqrt{109}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{109}}{2*21}=\frac{-32+2\sqrt{109}}{42}
$