21x^2+26x=0

Solution for 21x^2+26x=0 equation:

21x^2+26x=0

a = 21; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·21·0
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-26}{2*21}=\frac{-52}{42}
=-1+5/21
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*21}=\frac{0}{42}
=0
$