21x^2=11x+40

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

21x^2=11x+40

We move all terms to the left:

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

We get rid of parentheses

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