14x^2=21x-7

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

14x^2=21x-7

We move all terms to the left:

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

We get rid of parentheses

14x^2-21x+7=0

a = 14; b = -21; c = +7;
Δ = b2-4ac
Δ = -212-4·14·7
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*14}=\frac{14}{28}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+7}{2*14}=\frac{28}{28}
=1
$