21x^2-1050x+50=0

Solution for 21x^2-1050x+50=0 equation:

21x^2-1050x+50=0

a = 21; b = -1050; c = +50;
Δ = b2-4ac
Δ = -10502-4·21·50
Δ = 1098300
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:

$\sqrt{\Delta}=\sqrt{1098300}=\sqrt{100*10983}=\sqrt{100}*\sqrt{10983}=10\sqrt{10983}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1050)-10\sqrt{10983}}{2*21}=\frac{1050-10\sqrt{10983}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1050)+10\sqrt{10983}}{2*21}=\frac{1050+10\sqrt{10983}}{42}
$