21x^2-3x+49x-7=0

Solution for 21x^2-3x+49x-7=0 equation:

21x^2-3x+49x-7=0

We add all the numbers together, and all the variables

21x^2+46x-7=0

a = 21; b = 46; c = -7;
Δ = b2-4ac
Δ = 462-4·21·(-7)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-52}{2*21}=\frac{-98}{42}
=-2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+52}{2*21}=\frac{6}{42}
=1/7
$