21x^2=23x+20

Solution for 21x^2=23x+20 equation:

21x^2=23x+20

We move all terms to the left:

21x^2-(23x+20)=0

We get rid of parentheses

21x^2-23x-20=0

a = 21; b = -23; c = -20;
Δ = b2-4ac
Δ = -232-4·21·(-20)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-47}{2*21}=\frac{-24}{42}
=-4/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+47}{2*21}=\frac{70}{42}
=1+2/3
$