(21x^2+14x)=0

Solution for (21x^2+14x)=0 equation:

(21x^2+14x)=0

We get rid of parentheses

21x^2+14x=0

a = 21; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·21·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*21}=\frac{-28}{42}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*21}=\frac{0}{42}
=0
$