21x^2-25x+10=4

Solution for 21x^2-25x+10=4 equation:

21x^2-25x+10=4

We move all terms to the left:

21x^2-25x+10-(4)=0

We add all the numbers together, and all the variables

21x^2-25x+6=0

a = 21; b = -25; c = +6;
Δ = b2-4ac
Δ = -252-4·21·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-11}{2*21}=\frac{14}{42}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+11}{2*21}=\frac{36}{42}
=6/7
$