5x^2-20=21x

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

5x^2-20=21x

We move all terms to the left:

5x^2-20-(21x)=0

a = 5; b = -21; c = -20;
Δ = b2-4ac
Δ = -212-4·5·(-20)
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*5}=\frac{-8}{10}
=-4/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+29}{2*5}=\frac{50}{10}
=5
$