21x^2=5+32x

Solution for 21x^2=5+32x equation:

21x^2=5+32x

We move all terms to the left:

21x^2-(5+32x)=0

We add all the numbers together, and all the variables

21x^2-(32x+5)=0

We get rid of parentheses

21x^2-32x-5=0

a = 21; b = -32; c = -5;
Δ = b2-4ac
Δ = -322-4·21·(-5)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-38}{2*21}=\frac{-6}{42}
=-1/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+38}{2*21}=\frac{70}{42}
=1+2/3
$