21x^2=4x=25

Solution for 21x^2=4x=25 equation:

21x^2=4x=25

We move all terms to the left:

21x^2-(4x)=0

a = 21; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·21·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*21}=\frac{0}{42}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*21}=\frac{8}{42}
=4/21
$