21x^2-5x-35=3x^2-4x

Solution for 21x^2-5x-35=3x^2-4x equation:

21x^2-5x-35=3x^2-4x

We move all terms to the left:

21x^2-5x-35-(3x^2-4x)=0

We get rid of parentheses

21x^2-3x^2-5x+4x-35=0

We add all the numbers together, and all the variables

18x^2-1x-35=0

a = 18; b = -1; c = -35;
Δ = b2-4ac
Δ = -12-4·18·(-35)
Δ = 2521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{2521}}{2*18}=\frac{1-\sqrt{2521}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{2521}}{2*18}=\frac{1+\sqrt{2521}}{36}
$