20x^2+5=29x

Solution for 20x^2+5=29x equation:

20x^2+5=29x

We move all terms to the left:

20x^2+5-(29x)=0

a = 20; b = -29; c = +5;
Δ = b2-4ac
Δ = -292-4·20·5
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-21}{2*20}=\frac{8}{40}
=1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+21}{2*20}=\frac{50}{40}
=1+1/4
$