20x^2-5-21x=0

Solution for 20x^2-5-21x=0 equation:

20x^2-5-21x=0

a = 20; b = -21; c = -5;
Δ = b2-4ac
Δ = -212-4·20·(-5)
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{-(-21)-29}{2*20}=\frac{-8}{40}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+29}{2*20}=\frac{50}{40}
=1+1/4
$