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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

18x^2+4x-35=0

a = 18; b = 4; c = -35;
Δ = b2-4ac
Δ = 42-4·18·(-35)
Δ = 2536
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:

$\sqrt{\Delta}=\sqrt{2536}=\sqrt{4*634}=\sqrt{4}*\sqrt{634}=2\sqrt{634}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{634}}{2*18}=\frac{-4-2\sqrt{634}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{634}}{2*18}=\frac{-4+2\sqrt{634}}{36}
$