21x^2=10-x

Solution for 21x^2=10-x equation:

21x^2=10-x

We move all terms to the left:

21x^2-(10-x)=0

We add all the numbers together, and all the variables

21x^2-(-1x+10)=0

We get rid of parentheses

21x^2+1x-10=0

We add all the numbers together, and all the variables

21x^2+x-10=0

a = 21; b = 1; c = -10;
Δ = b2-4ac
Δ = 12-4·21·(-10)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-29}{2*21}=\frac{-30}{42}
=-5/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+29}{2*21}=\frac{28}{42}
=2/3
$