3x^2+21x+3=7x

Solution for 3x^2+21x+3=7x equation:

3x^2+21x+3=7x

We move all terms to the left:

3x^2+21x+3-(7x)=0

We add all the numbers together, and all the variables

3x^2+14x+3=0

a = 3; b = 14; c = +3;
Δ = b2-4ac
Δ = 142-4·3·3
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{10}}{2*3}=\frac{-14-4\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{10}}{2*3}=\frac{-14+4\sqrt{10}}{6}
$