21x^2-28x+7=0

Solution for 21x^2-28x+7=0 equation:

21x^2-28x+7=0

a = 21; b = -28; c = +7;
Δ = b2-4ac
Δ = -282-4·21·7
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-14}{2*21}=\frac{14}{42}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+14}{2*21}=\frac{42}{42}
=1
$