16x^2=21x

Solution for 16x^2=21x equation:

16x^2=21x

We move all terms to the left:

16x^2-(21x)=0

a = 16; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·16·0
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*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*16}=\frac{42}{32}
=1+5/16
$