21x^2-26+10=0

Solution for 21x^2-26+10=0 equation:

21x^2-26+10=0

We add all the numbers together, and all the variables

21x^2-16=0

a = 21; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·21·(-16)
Δ = 1344
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:

$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{21}}{2*21}=\frac{0-8\sqrt{21}}{42}
=-\frac{8\sqrt{21}}{42}
=-\frac{4\sqrt{21}}{21}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{21}}{2*21}=\frac{0+8\sqrt{21}}{42}
=\frac{8\sqrt{21}}{42}
=\frac{4\sqrt{21}}{21}
$