21x-2x^2=40

Solution for 21x-2x^2=40 equation:

21x-2x^2=40

We move all terms to the left:

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

a = -2; b = 21; c = -40;
Δ = b2-4ac
Δ = 212-4·(-2)·(-40)
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{-(21)-11}{2*-2}=\frac{-32}{-4}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*-2}=\frac{-10}{-4}
=2+1/2
$