X+10x^2=21

Solution for X+10x^2=21 equation:

X+10X^2=21

We move all terms to the left:

X+10X^2-(21)=0

a = 10; b = 1; c = -21;
Δ = b2-4ac
Δ = 12-4·10·(-21)
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*10}=\frac{-30}{20}
=-1+1/2
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+29}{2*10}=\frac{28}{20}
=1+2/5
$