7x^2+14=21x

Solution for 7x^2+14=21x equation:

7x^2+14=21x

We move all terms to the left:

7x^2+14-(21x)=0

a = 7; b = -21; c = +14;
Δ = b2-4ac
Δ = -212-4·7·14
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-7}{2*7}=\frac{14}{14}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+7}{2*7}=\frac{28}{14}
=2
$