21x^2-4x+6=7

Solution for 21x^2-4x+6=7 equation:

21x^2-4x+6=7

We move all terms to the left:

21x^2-4x+6-(7)=0

We add all the numbers together, and all the variables

21x^2-4x-1=0

a = 21; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·21·(-1)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*21}=\frac{-6}{42}
=-1/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*21}=\frac{14}{42}
=1/3
$