21x^2-62x+45=0

Solution for 21x^2-62x+45=0 equation:

21x^2-62x+45=0

a = 21; b = -62; c = +45;
Δ = b2-4ac
Δ = -622-4·21·45
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-8}{2*21}=\frac{54}{42}
=1+2/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+8}{2*21}=\frac{70}{42}
=1+2/3
$