21x^2+10x-20=0

Solution for 21x^2+10x-20=0 equation:

21x^2+10x-20=0

a = 21; b = 10; c = -20;
Δ = b2-4ac
Δ = 102-4·21·(-20)
Δ = 1780
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:

$\sqrt{\Delta}=\sqrt{1780}=\sqrt{4*445}=\sqrt{4}*\sqrt{445}=2\sqrt{445}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{445}}{2*21}=\frac{-10-2\sqrt{445}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{445}}{2*21}=\frac{-10+2\sqrt{445}}{42}
$