21x^2+9x=14x

Solution for 21x^2+9x=14x equation:

21x^2+9x=14x

We move all terms to the left:

21x^2+9x-(14x)=0

We add all the numbers together, and all the variables

21x^2-5x=0

a = 21; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·21·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*21}=\frac{0}{42}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*21}=\frac{10}{42}
=5/21
$