11x+40=21x^2

Solution for 11x+40=21x^2 equation:

11x+40=21x^2

We move all terms to the left:

11x+40-(21x^2)=0

determiningTheFunctionDomain
-21x^2+11x+40=0

a = -21; b = 11; c = +40;
Δ = b2-4ac
Δ = 112-4·(-21)·40
Δ = 3481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3481}=59$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-59}{2*-21}=\frac{-70}{-42}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+59}{2*-21}=\frac{48}{-42}
=-1+1/7
$