21x^2=16x+3

Solution for 21x^2=16x+3 equation:

21x^2=16x+3

We move all terms to the left:

21x^2-(16x+3)=0

We get rid of parentheses

21x^2-16x-3=0

a = 21; b = -16; c = -3;
Δ = b2-4ac
Δ = -162-4·21·(-3)
Δ = 508
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:

$\sqrt{\Delta}=\sqrt{508}=\sqrt{4*127}=\sqrt{4}*\sqrt{127}=2\sqrt{127}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{127}}{2*21}=\frac{16-2\sqrt{127}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{127}}{2*21}=\frac{16+2\sqrt{127}}{42}
$