28x^2=21x

Solution for 28x^2=21x equation:

28x^2=21x

We move all terms to the left:

28x^2-(21x)=0

a = 28; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·28·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*28}=\frac{0}{56}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*28}=\frac{42}{56}
=3/4
$